Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.8

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.8

Question 1. Find the principal solution and general solutions of the following
(i) \( \sin \theta = -\frac{1}{\sqrt{2}} \)
(ii) \( \cot \theta = \sqrt{3} \)
(iii) \( \tan \theta = -\frac{1}{\sqrt{3}} \)
Answer:
(i) Given \( \sin \theta = -\frac{1}{\sqrt{2}} \).
We know that the principal value of \( \sin \theta \) lies in the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
Since \( \sin \theta \) is negative, the principal solution lies in the IV quadrant.
We know that \( \sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \).
So, \( \sin \theta = -\sin \left(\frac{\pi}{4}\right) \)
\( \sin \theta = \sin \left(-\frac{\pi}{4}\right) \)
Hence, \( \theta = -\frac{\pi}{4} \) is the principal solution.
The general solution for \( \sin \theta = \sin \alpha \) is \( \theta = n\pi + (-1)^n \alpha \), where \( n \in Z \).
Substituting \( \alpha = -\frac{\pi}{4} \):
\( \theta = n\pi + (-1)^n \left(-\frac{\pi}{4}\right) \)
\( \theta = n\pi + (-1)^{n+1} \frac{\pi}{4}, n \in Z \)
(ii) Given \( \cot \theta = \sqrt{3} \).
We know that \( \tan \theta = \frac{1}{\cot \theta} \).
So, \( \tan \theta = \frac{1}{\sqrt{3}} \).
We know that the principal value of \( \tan \theta \) lies in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
Since \( \tan \theta = \frac{1}{\sqrt{3}} > 0 \), the principal solution lies in the I quadrant.
We know that \( \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \).
So, \( \tan \theta = \tan \left(\frac{\pi}{6}\right) \)
Hence, \( \theta = \frac{\pi}{6} \) is the principal solution.
The general solution for \( \tan \theta = \tan \alpha \) is \( \theta = n\pi + \alpha \), where \( n \in Z \).
Substituting \( \alpha = \frac{\pi}{6} \):
\( \theta = n\pi + \frac{\pi}{6}, n \in Z \)
(iii) Given \( \tan \theta = -\frac{1}{\sqrt{3}} \).
We know that the principal value of \( \tan \theta \) lies in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
Since \( \tan \theta = -\frac{1}{\sqrt{3}} < 0 \), the principal solution lies in the IV quadrant.
We know that \( \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \).
So, \( \tan \theta = -\tan \left(\frac{\pi}{6}\right) \)
\( \tan \theta = \tan \left(-\frac{\pi}{6}\right) \)
Hence, \( \theta = -\frac{\pi}{6} \) is the principal solution.
The general solution for \( \tan \theta = \tan \alpha \) is \( \theta = n\pi + \alpha \), where \( n \in Z \).
Substituting \( \alpha = -\frac{\pi}{6} \):
\( \theta = n\pi - \frac{\pi}{6}, n \in Z \)
In simple words: For each part, we first find the main angle (principal solution) that fits the given trigonometric value and is within the allowed range for that function. Then, we use the general solution formulas for sine, cosine, or tangent to find all possible angles. We remember that angles can repeat in cycles.

🎯 Exam Tip: Always specify the quadrant for principal solutions based on the sign of the trigonometric ratio. This helps in choosing the correct angle within the principal range.

Question 2. Solve the following equations for which solutions lies in the interval \( 0^\circ \le \theta < 360^\circ \)
(i) \( \sin^4 x = \sin^2 x \)
Answer:
(i) Given: \( \sin^4 x = \sin^2 x \)
Rearrange the equation: \( \sin^4 x - \sin^2 x = 0 \)
Factor out \( \sin^2 x \): \( \sin^2 x (\sin^2 x - 1) = 0 \)
Use the identity \( \sin^2 x - 1 = -\cos^2 x \):
\( \sin^2 x (-\cos^2 x) = 0 \)
\( -\sin^2 x \cos^2 x = 0 \)
\( (\sin x \cos x)^2 = 0 \)
Use the identity \( \sin 2x = 2 \sin x \cos x \), so \( \sin x \cos x = \frac{1}{2} \sin 2x \):
\( \left(\frac{1}{2} \sin 2x\right)^2 = 0 \)
\( \frac{1}{4} \sin^2 2x = 0 \)
\( \sin^2 2x = 0 \)
Taking the square root: \( \sin 2x = 0 \)
The general solution for \( \sin \phi = 0 \) is \( \phi = n\pi \), where \( n \in Z \).
So, \( 2x = n\pi \)
\( x = \frac{n\pi}{2}, n \in Z \)
Now, we find the values of \( x \) within the interval \( 0^\circ \le x < 360^\circ \) (or \( [0, 2\pi) \) in radians):
When \( n = 0, x = \frac{0 \cdot \pi}{2} = 0^\circ \). This value is in the interval.
When \( n = 1, x = \frac{1 \cdot \pi}{2} = \frac{\pi}{2} \) (or \( 90^\circ \)). This value is in the interval.
When \( n = 2, x = \frac{2 \cdot \pi}{2} = \pi \) (or \( 180^\circ \)). This value is in the interval.
When \( n = 3, x = \frac{3 \cdot \pi}{2} = \frac{3\pi}{2} \) (or \( 270^\circ \)). This value is in the interval.
When \( n = 4, x = \frac{4 \cdot \pi}{2} = 2\pi \) (or \( 360^\circ \)). This value is not in the interval because the upper bound is exclusive.
Therefore, the values of \( x \) are \( 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \).
In simple words: We changed the equation into a simpler form using trigonometric rules. We found that the sine of \( 2x \) must be zero. This means \( 2x \) can be any multiple of \( \pi \). Then, we divided by 2 to find \( x \). Finally, we listed all the angles for \( x \) that are between \( 0^\circ \) and just under \( 360^\circ \).

🎯 Exam Tip: When working with trigonometric equations, always simplify using identities first. Remember to check the specified interval for solutions and list all values within that range.

Question 2. Solve the following equations for which solutions lies in the interval \( 0^\circ \le \theta < 360^\circ \)
(ii) \( 2 \cos^2 x + 1 = -3 \cos x \)
Answer:
(ii) Given: \( 2 \cos^2 x + 1 = -3 \cos x \)
Rearrange the equation into a quadratic form by moving all terms to one side:
\( 2 \cos^2 x + 3 \cos x + 1 = 0 \)
Factor the quadratic equation by splitting the middle term:
\( 2 \cos^2 x + 2 \cos x + \cos x + 1 = 0 \)
Factor by grouping:
\( 2 \cos x (\cos x + 1) + 1 (\cos x + 1) = 0 \)
\( (2 \cos x + 1)(\cos x + 1) = 0 \)
This gives two possible conditions:
\( 2 \cos x + 1 = 0 \) or \( \cos x + 1 = 0 \)
So, \( \cos x = -\frac{1}{2} \) or \( \cos x = -1 \)

Case 1: \( \cos x = -\frac{1}{2} \)
We know that \( \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \). Since \( \cos x \) is negative, \( x \) lies in the second or third quadrant.
In the second quadrant, \( x = \pi - \frac{\pi}{3} = \frac{3\pi - \pi}{3} = \frac{2\pi}{3} \) (or \( 120^\circ \)). This is in the interval \( [0^\circ, 360^\circ) \).
In the third quadrant, \( x = \pi + \frac{\pi}{3} = \frac{3\pi + \pi}{3} = \frac{4\pi}{3} \) (or \( 240^\circ \)). This is in the interval \( [0^\circ, 360^\circ) \).

Case 2: \( \cos x = -1 \)
We know that \( x = \pi \) (or \( 180^\circ \)). This is in the interval \( [0^\circ, 360^\circ) \).
All values are within the interval. So, the required solutions are \( x = \frac{2\pi}{3}, \frac{4\pi}{3}, \pi \).
In simple words: We turned the equation into a simple quadratic equation using \( \cos x \). Then, we solved it to find two possible values for \( \cos x \): \( -\frac{1}{2} \) and \( -1 \). We then found all the angles that give these cosine values within the range of \( 0^\circ \) to \( 360^\circ \). These values show where the cosine function meets the required conditions.

🎯 Exam Tip: For quadratic trigonometric equations, treat the trigonometric function (like \( \cos x \)) as a variable. Factor the quadratic, then solve for each resulting simple trigonometric equation. Remember to find solutions in all relevant quadrants for the given range.

Question 2. Solve the following equations for which solutions lies in the interval \( 0^\circ \le \theta < 360^\circ \)
(iii) \( 2 \sin^2 x + 1 = 3 \sin x \)
Answer:
(iii) Given: \( 2 \sin^2 x + 1 = 3 \sin x \)
Rearrange the equation into a quadratic form by moving all terms to one side:
\( 2 \sin^2 x - 3 \sin x + 1 = 0 \)
Factor the quadratic equation by splitting the middle term:
\( 2 \sin^2 x - 2 \sin x - \sin x + 1 = 0 \)
Factor by grouping:
\( 2 \sin x (\sin x - 1) - 1 (\sin x - 1) = 0 \)
\( (2 \sin x - 1)(\sin x - 1) = 0 \)
This gives two possible conditions:
\( 2 \sin x - 1 = 0 \) or \( \sin x - 1 = 0 \)
So, \( \sin x = \frac{1}{2} \) or \( \sin x = 1 \)

Case 1: \( \sin x = \frac{1}{2} \)
We know that \( \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \). Since \( \sin x \) is positive, \( x \) lies in the first or second quadrant.
In the first quadrant, \( x = \frac{\pi}{6} \) (or \( 30^\circ \)). This is in the interval \( [0^\circ, 360^\circ) \).
In the second quadrant, \( x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \) (or \( 150^\circ \)). This is in the interval \( [0^\circ, 360^\circ) \).

Case 2: \( \sin x = 1 \)
We know that \( x = \frac{\pi}{2} \) (or \( 90^\circ \)). This is in the interval \( [0^\circ, 360^\circ) \).
All values are within the interval. So, the required solutions are \( x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{\pi}{2} \).
In simple words: We rearranged the equation into a quadratic form involving \( \sin x \) and then factored it. This gave us two possible values for \( \sin x \): \( \frac{1}{2} \) and \( 1 \). Finally, we found all the angles that correspond to these sine values within the specified range of \( 0^\circ \) to \( 360^\circ \). These are the angles where the sine function has those specific outputs.

🎯 Exam Tip: For solutions of \( \sin x = k \), always consider both the first quadrant angle and its supplementary angle ( \( \pi - \theta \) ) if \( k \) is positive, or relevant angles in third and fourth quadrants if \( k \) is negative, to cover all possibilities within a full cycle.

Question 2. Solve the following equations for which solutions lies in the interval \( 0^\circ \le \theta < 360^\circ \)
(iv) \( \cos^2 x = 1 - 3 \sin x \)
Answer:
(iv) Given: \( \cos^2 x = 1 - 3 \sin x \)
Use the identity \( \cos^2 x = 1 - \sin^2 x \):
\( 1 - \sin^2 x = 1 - 3 \sin x \)
Subtract 1 from both sides:
\( -\sin^2 x = -3 \sin x \)
Move all terms to one side and multiply by -1:
\( \sin^2 x - 3 \sin x = 0 \)
Factor out \( \sin x \):
\( \sin x (\sin x - 3) = 0 \)
This gives two possible conditions:
\( \sin x = 0 \) or \( \sin x - 3 = 0 \)

Case 1: \( \sin x = 0 \)
The general solution for \( \sin \phi = 0 \) is \( \phi = n\pi \), where \( n \in Z \).
We find the values of \( x \) within the interval \( 0^\circ \le x < 360^\circ \) (or \( [0, 2\pi) \) in radians):
When \( n = 0, x = 0 \cdot \pi = 0^\circ \). This value is in the interval.
When \( n = 1, x = 1 \cdot \pi = \pi \) (or \( 180^\circ \)). This value is in the interval.
When \( n = 2, x = 2 \cdot \pi = 2\pi \) (or \( 360^\circ \)). This value is not in the interval because the upper bound is exclusive.

Case 2: \( \sin x = 3 \)
This is not possible because the value of \( \sin x \) must always be between -1 and 1 (inclusive).
Therefore, the required solutions are \( x = 0, \pi \).
In simple words: We replaced \( \cos^2 x \) with \( 1 - \sin^2 x \) to get an equation with only \( \sin x \). Then, we solved it like a simple quadratic equation by factoring. We found that \( \sin x \) must be 0, because \( \sin x \) can never be 3. The angles where \( \sin x \) is 0, within \( 0^\circ \) and \( 360^\circ \), are \( 0^\circ \) and \( 180^\circ \).

🎯 Exam Tip: Always remember that the range of sine and cosine functions is \( [-1, 1] \). Any solution outside this range (like \( \sin x = 3 \)) is extraneous and must be discarded.

Question 3. Solve the following equations:
(i) \( \sin 5x - \sin x = \cos 3x \)
Answer:
(i) Given: \( \sin 5x - \sin x = \cos 3x \)
Use the sum-to-product formula for sine: \( \sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \)
Applying this to the left side:
\( 2 \cos \left(\frac{5x+x}{2}\right) \sin \left(\frac{5x-x}{2}\right) = \cos 3x \)
\( 2 \cos \left(\frac{6x}{2}\right) \sin \left(\frac{4x}{2}\right) = \cos 3x \)
\( 2 \cos 3x \sin 2x = \cos 3x \)
Move all terms to one side:
\( 2 \cos 3x \sin 2x - \cos 3x = 0 \)
Factor out \( \cos 3x \):
\( \cos 3x (2 \sin 2x - 1) = 0 \)
This gives two possible conditions:
\( \cos 3x = 0 \) or \( 2 \sin 2x - 1 = 0 \)

Case 1: \( \cos 3x = 0 \)
The general solution for \( \cos \phi = 0 \) is \( \phi = (2n + 1)\frac{\pi}{2} \), where \( n \in Z \).
So, \( 3x = (2n + 1)\frac{\pi}{2} \)
\( x = (2n + 1)\frac{\pi}{6}, n \in Z \)

Case 2: \( 2 \sin 2x - 1 = 0 \)
So, \( \sin 2x = \frac{1}{2} \)
We know that \( \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \).
The general solution for \( \sin \phi = \sin \alpha \) is \( \phi = m\pi + (-1)^m \alpha \), where \( m \in Z \).
So, \( 2x = m\pi + (-1)^m \frac{\pi}{6} \)
\( x = \frac{m\pi}{2} + (-1)^m \frac{\pi}{12}, m \in Z \)
In simple words: We used a special formula to change the left side of the equation from two sines to a product of sine and cosine. After simplifying, we factored the equation to get two separate parts: one where \( \cos 3x = 0 \) and another where \( \sin 2x = \frac{1}{2} \). We then found the general solutions for each of these simpler equations, which tell us all the possible angles that fit the conditions. These two sets of solutions together solve the original equation.

🎯 Exam Tip: When using sum-to-product or product-to-sum formulas, correctly identify A and B and perform the arithmetic for \( \frac{A+B}{2} \) and \( \frac{A-B}{2} \) carefully. Factoring after simplification is a common and effective strategy.

Question 3. Solve the following equations:
(ii) \( 2 \cos^2 \theta + 3 \sin \theta - 3 = 0 \)
Answer:
(ii) Given: \( 2 \cos^2 \theta + 3 \sin \theta - 3 = 0 \)
Use the identity \( \cos^2 \theta = 1 - \sin^2 \theta \) to express the equation only in terms of \( \sin \theta \):
\( 2(1 - \sin^2 \theta) + 3 \sin \theta - 3 = 0 \)
Distribute the 2:
\( 2 - 2 \sin^2 \theta + 3 \sin \theta - 3 = 0 \)
Combine constant terms:
\( -2 \sin^2 \theta + 3 \sin \theta - 1 = 0 \)
Multiply the entire equation by -1 to make the leading term positive:
\( 2 \sin^2 \theta - 3 \sin \theta + 1 = 0 \)
Factor the quadratic equation by splitting the middle term:
\( 2 \sin^2 \theta - 2 \sin \theta - \sin \theta + 1 = 0 \)
Factor by grouping:
\( 2 \sin \theta (\sin \theta - 1) - 1 (\sin \theta - 1) = 0 \)
\( (2 \sin \theta - 1)(\sin \theta - 1) = 0 \)
This gives two possible conditions:
\( 2 \sin \theta - 1 = 0 \) or \( \sin \theta - 1 = 0 \)
So, \( \sin \theta = \frac{1}{2} \) or \( \sin \theta = 1 \)

Case 1: \( \sin \theta = \frac{1}{2} \)
We know that \( \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \).
The general solution for \( \sin \phi = \sin \alpha \) is \( \phi = n\pi + (-1)^n \alpha \), where \( n \in Z \).
So, \( \theta = n\pi + (-1)^n \frac{\pi}{6}, n \in Z \)

Case 2: \( \sin \theta = 1 \)
We know that \( \sin \left(\frac{\pi}{2}\right) = 1 \).
The general solution for \( \sin \phi = \sin \alpha \) is \( \phi = n\pi + (-1)^n \alpha \), where \( n \in Z \).
So, \( \theta = n\pi + (-1)^n \frac{\pi}{2}, n \in Z \)
In simple words: We changed the equation so it only had \( \sin \theta \) using a common identity. This made it a quadratic equation, which we factored into two parts. From these parts, we found that \( \sin \theta \) must either be \( \frac{1}{2} \) or \( 1 \). Finally, we wrote down the general formulas for all angles \( \theta \) that make these sine values true, which covers all solutions.

🎯 Exam Tip: When an equation contains both sine and cosine squared terms, it's often best to convert everything to a single trigonometric function using identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) before attempting to solve.

Question 3. Solve the following equations:
(iii) \( \cos \theta + \cos 3\theta = 2 \cos 2\theta \)
Answer:
(iii) Given: \( \cos \theta + \cos 3\theta = 2 \cos 2\theta \)
Rearrange the left side: \( \cos 3\theta + \cos \theta = 2 \cos 2\theta \)
Use the sum-to-product formula for cosine: \( \cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) \)
Applying this to the left side:
\( 2 \cos \left(\frac{3\theta+\theta}{2}\right) \cos \left(\frac{3\theta-\theta}{2}\right) = 2 \cos 2\theta \)
\( 2 \cos \left(\frac{4\theta}{2}\right) \cos \left(\frac{2\theta}{2}\right) = 2 \cos 2\theta \)
\( 2 \cos 2\theta \cos \theta = 2 \cos 2\theta \)
Divide both sides by 2:
\( \cos 2\theta \cos \theta = \cos 2\theta \)
Move all terms to one side:
\( \cos 2\theta \cos \theta - \cos 2\theta = 0 \)
Factor out \( \cos 2\theta \):
\( \cos 2\theta (\cos \theta - 1) = 0 \)
This gives two possible conditions:
\( \cos 2\theta = 0 \) or \( \cos \theta - 1 = 0 \)

Case 1: \( \cos 2\theta = 0 \)
The general solution for \( \cos \phi = 0 \) is \( \phi = (2n + 1)\frac{\pi}{2} \), where \( n \in Z \).
So, \( 2\theta = (2n + 1)\frac{\pi}{2} \)
\( \theta = (2n + 1)\frac{\pi}{4}, n \in Z \)

Case 2: \( \cos \theta - 1 = 0 \)
So, \( \cos \theta = 1 \)
The general solution for \( \cos \phi = 1 \) is \( \phi = 2n\pi \), where \( n \in Z \).
So, \( \theta = 2n\pi, n \in Z \)
In simple words: We used a sum-to-product formula to simplify the left side of the equation. After some steps, we moved all terms to one side and factored out a common term, \( \cos 2\theta \). This led to two simpler equations: \( \cos 2\theta = 0 \) and \( \cos \theta = 1 \). We then found the general solutions for each of these. These two sets of solutions together cover all the angles that solve the original equation.

🎯 Exam Tip: Always remember to look for common factors once you've applied sum-to-product or product-to-sum formulas. Factoring allows you to break down a complex equation into simpler, solvable parts.

Question 3. Solve the following equations:
(iv) \( \sin \theta + \sin 3\theta + \sin 5\theta = 0 \)
Answer:
(iv) Given: \( \sin \theta + \sin 3\theta + \sin 5\theta = 0 \)
Rearrange and group terms: \( (\sin 5\theta + \sin \theta) + \sin 3\theta = 0 \)
Use the sum-to-product formula for sine: \( \sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) \)
Applying this to \( (\sin 5\theta + \sin \theta) \):
\( 2 \sin \left(\frac{5\theta+\theta}{2}\right) \cos \left(\frac{5\theta-\theta}{2}\right) + \sin 3\theta = 0 \)
\( 2 \sin \left(\frac{6\theta}{2}\right) \cos \left(\frac{4\theta}{2}\right) + \sin 3\theta = 0 \)
\( 2 \sin 3\theta \cos 2\theta + \sin 3\theta = 0 \)
Factor out \( \sin 3\theta \):
\( \sin 3\theta (2 \cos 2\theta + 1) = 0 \)
This gives two possible conditions:
\( \sin 3\theta = 0 \) or \( 2 \cos 2\theta + 1 = 0 \)

Case 1: \( \sin 3\theta = 0 \)
The general solution for \( \sin \phi = 0 \) is \( \phi = n\pi \), where \( n \in Z \).
So, \( 3\theta = n\pi \)
\( \theta = \frac{n\pi}{3}, n \in Z \)

Case 2: \( 2 \cos 2\theta + 1 = 0 \)
So, \( \cos 2\theta = -\frac{1}{2} \)
We know that \( \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \). Since \( \cos 2\theta \) is negative, \( 2\theta \) lies in the second or third quadrant. The reference angle is \( \frac{\pi}{3} \). In the second quadrant, \( \alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \).
The general solution for \( \cos \phi = \cos \alpha \) is \( \phi = 2n\pi \pm \alpha \), where \( n \in Z \).
So, \( 2\theta = 2n\pi \pm \frac{2\pi}{3} \)
Divide by 2:
\( \theta = n\pi \pm \frac{\pi}{3}, n \in Z \)
In simple words: We grouped terms and used a sine sum-to-product formula to simplify the equation. This allowed us to factor out \( \sin 3\theta \), creating two simpler equations: \( \sin 3\theta = 0 \) and \( 2 \cos 2\theta + 1 = 0 \). We then found all the general solutions for each of these, covering all angles where \( \sin 3\theta \) is zero or \( \cos 2\theta \) is \( -\frac{1}{2} \).

🎯 Exam Tip: When dealing with more than two terms, try grouping two terms that simplify nicely using sum-to-product formulas to create a common factor, as was done with \( \sin 5\theta + \sin \theta \).

Question 3. Solve the following equations:
(v) \( \sin 2\theta - \cos 2\theta - \sin \theta + \cos \theta = 0 \)
Answer:
(v) Given: \( \sin 2\theta - \cos 2\theta - \sin \theta + \cos \theta = 0 \)
Rearrange and group terms:
\( (\sin 2\theta - \sin \theta) - (\cos 2\theta - \cos \theta) = 0 \)
Use sum-to-product formulas:
For \( \sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \):
\( \sin 2\theta - \sin \theta = 2 \cos \left(\frac{2\theta+\theta}{2}\right) \sin \left(\frac{2\theta-\theta}{2}\right) = 2 \cos \frac{3\theta}{2} \sin \frac{\theta}{2} \)
For \( \cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \):
\( \cos 2\theta - \cos \theta = -2 \sin \left(\frac{2\theta+\theta}{2}\right) \sin \left(\frac{2\theta-\theta}{2}\right) = -2 \sin \frac{3\theta}{2} \sin \frac{\theta}{2} \)
Substitute these back into the equation:
\( 2 \cos \frac{3\theta}{2} \sin \frac{\theta}{2} - \left(-2 \sin \frac{3\theta}{2} \sin \frac{\theta}{2}\right) = 0 \)
\( 2 \cos \frac{3\theta}{2} \sin \frac{\theta}{2} + 2 \sin \frac{3\theta}{2} \sin \frac{\theta}{2} = 0 \)
Factor out \( 2 \sin \frac{\theta}{2} \):
\( 2 \sin \frac{\theta}{2} \left(\cos \frac{3\theta}{2} + \sin \frac{3\theta}{2}\right) = 0 \)
This gives two possible conditions:
\( \sin \frac{\theta}{2} = 0 \) or \( \cos \frac{3\theta}{2} + \sin \frac{3\theta}{2} = 0 \)

Case 1: \( \sin \frac{\theta}{2} = 0 \)
The general solution for \( \sin \phi = 0 \) is \( \phi = n\pi \), where \( n \in Z \).
So, \( \frac{\theta}{2} = n\pi \)
\( \theta = 2n\pi, n \in Z \)

Case 2: \( \cos \frac{3\theta}{2} + \sin \frac{3\theta}{2} = 0 \)
Divide by \( \cos \frac{3\theta}{2} \) (assuming \( \cos \frac{3\theta}{2} \neq 0 \)):
\( 1 + \tan \frac{3\theta}{2} = 0 \)
\( \tan \frac{3\theta}{2} = -1 \)
We know that \( \tan \left(-\frac{\pi}{4}\right) = -1 \).
The general solution for \( \tan \phi = \tan \alpha \) is \( \phi = n\pi + \alpha \), where \( n \in Z \).
So, \( \frac{3\theta}{2} = n\pi - \frac{\pi}{4} \)
Multiply by \( \frac{2}{3} \):
\( \theta = \frac{2n\pi}{3} - \frac{2\pi}{12} \)
\( \theta = \frac{2n\pi}{3} - \frac{\pi}{6}, n \in Z \)
Alternatively, using \( \alpha = \frac{3\pi}{4} \) (since \( \tan \frac{3\pi}{4} = -1 \)):
\( \frac{3\theta}{2} = n\pi + \frac{3\pi}{4} \)
\( \theta = \frac{2n\pi}{3} + \frac{6\pi}{12} \)
\( \theta = \frac{2n\pi}{3} + \frac{\pi}{2}, n \in Z \)
In simple words: We first grouped terms in the equation to use special trigonometric formulas that combine sines and cosines. This helped us simplify the equation and factor out \( 2 \sin \frac{\theta}{2} \). This gave us two easier equations to solve: one where \( \sin \frac{\theta}{2} \) is zero, and another where \( \tan \frac{3\theta}{2} \) is \( -1 \). We then found the general solutions for each part, which describe all possible angles that make the original equation true.

🎯 Exam Tip: When an equation has a mix of sine and cosine terms with different angles, try to use sum-to-product or double angle formulas to unify the angles or create common factors. Also, remember that \( \cos x + \sin x = 0 \) simplifies to \( \tan x = -1 \).

Question 3. Solve the following equations:
(vi) \( \sin \theta + \sqrt{3} \cos \theta = \sqrt{2} \)
Answer:
(vi) Given: \( \sin \theta + \sqrt{3} \cos \theta = \sqrt{2} \)
To solve this equation, we divide by \( R = \sqrt{a^2 + b^2} \), where \( a=1 \) and \( b=\sqrt{3} \).
\( R = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2 \)
Divide each term of the equation by 2:
\( \frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta = \frac{\sqrt{2}}{2} \)
We know that \( \cos \frac{\pi}{3} = \frac{1}{2} \) and \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \). Also, \( \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4} \).
Substitute these values into the equation:
\( \cos \frac{\pi}{3} \sin \theta + \sin \frac{\pi}{3} \cos \theta = \sin \frac{\pi}{4} \)
Use the trigonometric identity \( \sin(A+B) = \sin A \cos B + \cos A \sin B \):
\( \sin \left(\theta + \frac{\pi}{3}\right) = \sin \frac{\pi}{4} \)
The general solution for \( \sin \phi = \sin \alpha \) is \( \phi = n\pi + (-1)^n \alpha \), where \( n \in Z \).
So, \( \theta + \frac{\pi}{3} = n\pi + (-1)^n \frac{\pi}{4} \)
Subtract \( \frac{\pi}{3} \) from both sides to solve for \( \theta \):
\( \theta = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{3}, n \in Z \)
In simple words: We converted the left side of the equation into a single sine function using a standard method of dividing by \( \sqrt{a^2+b^2} \). This transformed the equation into a simple \( \sin(\theta + \text{angle}) = \text{value} \) form. Then, we used the general solution for sine to find all possible angles for \( \theta \) that satisfy the equation. This method helps to combine different trigonometric terms into one solvable form.

🎯 Exam Tip: Equations of the form \( a \sin \theta + b \cos \theta = c \) are best solved by converting them to \( R \sin(\theta+\alpha) = c \) or \( R \cos(\theta-\alpha) = c \). Remember \( R = \sqrt{a^2+b^2} \) and \( \alpha \) is found using \( \tan \alpha = \frac{b}{a} \).

Question 3. Solve the following equations for which solutions lies in the interval 0° ≤ θ < 360°
(vii) sin θ + √3 cos θ = 1
Answer:
We are given the equation \( \sin \theta + \sqrt{3} \cos \theta = 1 \).
First, we divide every term in the equation by 2. This helps us use standard trigonometric identities.
\( \frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta = \frac{1}{2} \)
We know that \( \cos \frac{\pi}{3} = \frac{1}{2} \) and \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \).
So, we can rewrite the equation as: \( \cos \frac{\pi}{3} \sin \theta + \sin \frac{\pi}{3} \cos \theta = \frac{1}{2} \)
This matches the identity \( \sin(A+B) = \sin A \cos B + \cos A \sin B \).
\( \sin \left( \theta + \frac{\pi}{3} \right) = \frac{1}{2} \)
We know that \( \sin \frac{\pi}{6} = \frac{1}{2} \).
So, \( \sin \left( \theta + \frac{\pi}{3} \right) = \sin \frac{\pi}{6} \)
The general solution for \( \sin x = \sin \alpha \) is \( x = n\pi + (-1)^n \alpha \), where \( n \in Z \).
Applying this, we get:
\( \theta + \frac{\pi}{3} = n\pi + (-1)^n \frac{\pi}{6} \)
\( \implies \theta = n\pi + (-1)^n \frac{\pi}{6} - \frac{\pi}{3} \), where \( n \in Z \)
Let's find the specific solutions by considering even and odd values of \( n \).
Case 1: If \( n \) is an even integer, let \( n = 2k \).
\( \theta = 2k\pi + (-1)^{2k} \frac{\pi}{6} - \frac{\pi}{3} \)
\( \implies \theta = 2k\pi + \frac{\pi}{6} - \frac{\pi}{3} \)
\( \implies \theta = 2k\pi + \frac{\pi - 2\pi}{6} \)
\( \implies \theta = 2k\pi - \frac{\pi}{6} \)
Case 2: If \( n \) is an odd integer, let \( n = 2k+1 \).
\( \theta = (2k+1)\pi + (-1)^{2k+1} \frac{\pi}{6} - \frac{\pi}{3} \)
\( \implies \theta = (2k+1)\pi - \frac{\pi}{6} - \frac{\pi}{3} \)
\( \implies \theta = 2k\pi + \pi - \frac{\pi}{6} - \frac{2\pi}{6} \)
\( \implies \theta = 2k\pi + \frac{6\pi - \pi - 2\pi}{6} \)
\( \implies \theta = 2k\pi + \frac{3\pi}{6} \)
\( \implies \theta = 2k\pi + \frac{\pi}{2} \)
So, the required solutions are \( \theta = 2n\pi - \frac{\pi}{6} \) and \( \theta = 2n\pi + \frac{\pi}{2} \), for \( n \in Z \).
In simple words: To solve this, we change the equation so it looks like a simple sine function. Then we use a general rule for sine to find all possible answers for \( \theta \). We check for both even and odd values of 'n' to find the two main forms of the solution.

🎯 Exam Tip: When an equation has both sine and cosine terms, try to convert it into a single trigonometric function (like sine or cosine) using appropriate identities, often by dividing by a constant like the hypotenuse of (1, √3).

Question 3. Solve the following equations for which solutions lies in the interval 0° ≤ θ < 360°
(viii) cot θ + cosec θ = √3
Answer:
We are given the equation \( \cot \theta + \operatorname{cosec} \theta = \sqrt{3} \).
First, we rewrite cotangent and cosecant in terms of sine and cosine.
\( \frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta} = \sqrt{3} \)
Combine the fractions on the left side:
\( \frac{\cos \theta + 1}{\sin \theta} = \sqrt{3} \)
Multiply both sides by \( \sin \theta \):
\( 1 + \cos \theta = \sqrt{3} \sin \theta \)
Now, we rearrange the equation to prepare it for solving:
\( \sqrt{3} \sin \theta - \cos \theta = 1 \)
To solve this, we divide every term by 2.
\( \frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta = \frac{1}{2} \)
We know that \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \) and \( \sin \frac{\pi}{6} = \frac{1}{2} \).
So, we can rewrite the equation as: \( \sin \theta \cos \frac{\pi}{6} - \cos \theta \sin \frac{\pi}{6} = \frac{1}{2} \)
This matches the identity \( \sin(A-B) = \sin A \cos B - \cos A \sin B \).
\( \sin \left( \theta - \frac{\pi}{6} \right) = \frac{1}{2} \)
We know that \( \sin \frac{\pi}{6} = \frac{1}{2} \).
So, \( \sin \left( \theta - \frac{\pi}{6} \right) = \sin \frac{\pi}{6} \)
The general solution for \( \sin x = \sin \alpha \) is \( x = n\pi + (-1)^n \alpha \), where \( n \in Z \).
Applying this, we get:
\( \theta - \frac{\pi}{6} = n\pi + (-1)^n \frac{\pi}{6} \)
\( \implies \theta = n\pi + (-1)^n \frac{\pi}{6} + \frac{\pi}{6} \), where \( n \in Z \)
Let's find the specific solutions by considering even and odd values of \( n \).
Case 1: If \( n \) is an even integer, let \( n = 2k \).
\( \theta = 2k\pi + (-1)^{2k} \frac{\pi}{6} + \frac{\pi}{6} \)
\( \implies \theta = 2k\pi + \frac{\pi}{6} + \frac{\pi}{6} \)
\( \implies \theta = 2k\pi + \frac{2\pi}{6} \)
\( \implies \theta = 2k\pi + \frac{\pi}{3} \)
Case 2: If \( n \) is an odd integer, let \( n = 2k+1 \).
\( \theta = (2k+1)\pi + (-1)^{2k+1} \frac{\pi}{6} + \frac{\pi}{6} \)
\( \implies \theta = (2k+1)\pi - \frac{\pi}{6} + \frac{\pi}{6} \)
\( \implies \theta = (2k+1)\pi \)
However, the original equation has \( \cot \theta \) and \( \operatorname{cosec} \theta \), which means \( \sin \theta \) cannot be zero. If \( \theta = (2k+1)\pi \), then \( \sin \theta = 0 \), which is not allowed. Therefore, this case is not a valid solution.
So, the only required solution is \( \theta = 2n\pi + \frac{\pi}{3} \), for \( n \in Z \).
In simple words: We change the cotangent and cosecant parts into sine and cosine. After some rearranging, we get a simpler sine equation. Then we use the general rule for sine to find the answers, making sure to avoid any answers that would make the original problem undefined.

🎯 Exam Tip: Always remember to check the domain of the original equation when finding general solutions, especially for functions like cotangent and cosecant, to exclude values where the denominator is zero.

Question 3. Solve the following equations for which solutions lies in the interval 0° ≤ θ < 360°
(ix) tan θ + tan \( \left(\theta+\frac{\pi}{3}\right) \) + tan \( \left(\theta+\frac{2 \pi}{3}\right) = \sqrt{3} \)
Answer:
We are given the equation \( \tan \theta + \tan \left(\theta+\frac{\pi}{3}\right) + \tan \left(\theta+\frac{2 \pi}{3}\right) = \sqrt{3} \).
We use the identity \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \).
Also, we know that \( \tan \frac{\pi}{3} = \sqrt{3} \) and \( \tan \frac{2\pi}{3} = -\sqrt{3} \).
Substitute these values into the equation:
\( \tan \theta + \frac{\tan \theta + \tan \frac{\pi}{3}}{1 - \tan \theta \tan \frac{\pi}{3}} + \frac{\tan \theta + \tan \frac{2\pi}{3}}{1 - \tan \theta \tan \frac{2\pi}{3}} = \sqrt{3} \)
\( \tan \theta + \frac{\tan \theta + \sqrt{3}}{1 - \sqrt{3} \tan \theta} + \frac{\tan \theta - \sqrt{3}}{1 - (-\sqrt{3}) \tan \theta} = \sqrt{3} \)
\( \tan \theta + \frac{\tan \theta + \sqrt{3}}{1 - \sqrt{3} \tan \theta} + \frac{\tan \theta - \sqrt{3}}{1 + \sqrt{3} \tan \theta} = \sqrt{3} \)
Combine the two fractions by finding a common denominator:
\( \tan \theta + \frac{(\tan \theta + \sqrt{3})(1 + \sqrt{3} \tan \theta) + (\tan \theta - \sqrt{3})(1 - \sqrt{3} \tan \theta)}{(1 - \sqrt{3} \tan \theta)(1 + \sqrt{3} \tan \theta)} = \sqrt{3} \)
Expand the terms in the numerator:
\( (\tan \theta + \sqrt{3} \tan^2 \theta + \sqrt{3} + 3 \tan \theta) + (\tan \theta - \sqrt{3} \tan^2 \theta - \sqrt{3} + 3 \tan \theta) \)
\( = \tan \theta + \sqrt{3} \tan^2 \theta + \sqrt{3} + 3 \tan \theta + \tan \theta - \sqrt{3} \tan^2 \theta - \sqrt{3} + 3 \tan \theta \)
Combine like terms:
\( = (1+3+1+3)\tan \theta + (\sqrt{3}-\sqrt{3})\tan^2 \theta + (\sqrt{3}-\sqrt{3}) \)
\( = 8 \tan \theta \)
The denominator is \( (1 - \sqrt{3} \tan \theta)(1 + \sqrt{3} \tan \theta) = 1^2 - (\sqrt{3} \tan \theta)^2 = 1 - 3 \tan^2 \theta \).
So, the equation becomes:
\( \tan \theta + \frac{8 \tan \theta}{1 - 3 \tan^2 \theta} = \sqrt{3} \)
Bring all terms to a common denominator:
\( \frac{\tan \theta (1 - 3 \tan^2 \theta) + 8 \tan \theta}{1 - 3 \tan^2 \theta} = \sqrt{3} \)
\( \frac{\tan \theta - 3 \tan^3 \theta + 8 \tan \theta}{1 - 3 \tan^2 \theta} = \sqrt{3} \)
\( \frac{9 \tan \theta - 3 \tan^3 \theta}{1 - 3 \tan^2 \theta} = \sqrt{3} \)
Factor out 3 from the numerator:
\( 3 \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} = \sqrt{3} \)
We know the triple angle identity \( \tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \).
So, the equation simplifies to:
\( 3 \tan 3\theta = \sqrt{3} \)
\( \tan 3\theta = \frac{\sqrt{3}}{3} \)
\( \tan 3\theta = \frac{1}{\sqrt{3}} \)
We know that \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \).
So, \( \tan 3\theta = \tan \frac{\pi}{6} \)
The general solution for \( \tan x = \tan \alpha \) is \( x = n\pi + \alpha \), where \( n \in Z \).
Applying this, we get:
\( 3\theta = n\pi + \frac{\pi}{6} \)
Divide by 3 to find \( \theta \):
\( \theta = \frac{n\pi}{3} + \frac{\pi}{18} \), where \( n \in Z \).
In simple words: This problem asks us to solve an equation with three tangent terms. We use a special formula for tangent of (A+B) and then simplify the whole thing down to a simpler tangent equation for \( 3\theta \). Finally, we use the general rule for tangent to find all the possible values of \( \theta \).

🎯 Exam Tip: Recognizing and applying triple angle identities like \( \tan 3\theta \) can significantly simplify complex trigonometric equations, so always look for opportunities to use them.

Question 3. Solve the following equations for which solutions lies in the interval 0° ≤ θ < 360°
(x) cos \( 2\theta = \frac{\sqrt{5}+1}{4} \)
Answer:
We are given the equation \( \cos 2\theta = \frac{\sqrt{5}+1}{4} \).
We know a special trigonometric value: \( \cos 36^\circ = \cos \frac{\pi}{5} = \frac{\sqrt{5}+1}{4} \). This is a useful value to remember for many problems.
Comparing the given equation with this known value:
\( \cos 2\theta = \cos \frac{\pi}{5} \)
The general solution for \( \cos x = \cos \alpha \) is \( x = 2n\pi \pm \alpha \), where \( n \in Z \).
Applying this to our equation:
\( 2\theta = 2n\pi \pm \frac{\pi}{5} \)
To find \( \theta \), we divide both sides by 2:
\( \theta = \frac{2n\pi}{2} \pm \frac{\pi}{5 \times 2} \)
\( \theta = n\pi \pm \frac{\pi}{10} \), where \( n \in Z \).
In simple words: We are given a cosine equation with a specific number on the right side. We recognize that number as the cosine of a known angle, \( 36^\circ \). Once we replace the number with \( \cos 36^\circ \), we can use the general rule for cosine equations to find all possible values of \( \theta \).

🎯 Exam Tip: Memorizing common exact trigonometric values for angles like \( 15^\circ, 18^\circ, 36^\circ, 54^\circ, 72^\circ \) can greatly speed up solving problems that feature these specific values.

Question 3. Solve the following equations for which solutions lies in the interval 0° ≤ θ < 360°
(xi) \( 2\cos^2 x – 7 \cos x + 3 = 0 \)
Answer:
We are given the equation \( 2\cos^2 x – 7 \cos x + 3 = 0 \).
This is a quadratic equation in terms of \( \cos x \). Let \( y = \cos x \).
Then the equation becomes:
\( 2y^2 - 7y + 3 = 0 \)
We can solve this quadratic equation by factorization.
Break down the middle term \( -7y \) into \( -6y \) and \( -y \):
\( 2y^2 - 6y - y + 3 = 0 \)
Group the terms and factor out common factors:
\( 2y(y - 3) - 1(y - 3) = 0 \)
Factor out the common binomial \( (y - 3) \):
\( (2y - 1)(y - 3) = 0 \)
For the product of two terms to be zero, at least one of them must be zero.
So, either \( 2y - 1 = 0 \) or \( y - 3 = 0 \).
Case 1: \( 2y - 1 = 0 \)
\( 2y = 1 \)
\( y = \frac{1}{2} \)
Substitute back \( y = \cos x \):
\( \cos x = \frac{1}{2} \)
Case 2: \( y - 3 = 0 \)
\( y = 3 \)
Substitute back \( y = \cos x \):
\( \cos x = 3 \)
However, the range of the cosine function is \( -1 \le \cos x \le 1 \). This means that \( \cos x \) can never be 3.
So, the solution \( \cos x = 3 \) is not possible.
We only need to consider \( \cos x = \frac{1}{2} \).
We know that \( \cos \frac{\pi}{3} = \frac{1}{2} \).
So, \( \cos x = \cos \frac{\pi}{3} \)
The general solution for \( \cos x = \cos \alpha \) is \( x = 2n\pi \pm \alpha \), where \( n \in Z \).
Applying this, we get:
\( x = 2n\pi \pm \frac{\pi}{3} \), where \( n \in Z \).
In simple words: This is a math puzzle where we need to find the value of 'x' when 'cos x' is part of a square equation. We first treat 'cos x' like a regular letter in the equation and solve for it. Once we find the values for 'cos x', we check if they are possible (because 'cos x' can only be between -1 and 1). Finally, we use the general rule for cosine to find the complete answer for 'x'.

🎯 Exam Tip: When solving trigonometric equations that are quadratic in form, always remember to check if the solutions for the trigonometric function (like \( \cos x \) or \( \sin x \)) fall within their valid range of \([-1, 1]\). Solutions outside this range must be discarded.