OP Malhotra Class 11 Maths Solutions Chapter 6 Trigonometric Equations Chapter Test

Question 1. Find the general value of \( \theta \) which satisfies the equation \( \sin^2 \theta = \frac { 3 }{ 4 } \).
Answer:
We are given the equation: \( \sin^2 \theta = \frac{3}{4} \)
We can rewrite \( \frac{3}{4} \) as \( \left(\frac{\sqrt{3}}{2}\right)^2 \).
So, \( \sin^2 \theta = \left(\frac{\sqrt{3}}{2}\right)^2 \)
We know that \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \).
Thus, \( \sin^2 \theta = \sin^2 \frac{\pi}{3} \)
When \( \sin^2 \theta = \sin^2 \alpha \), the general solution for \( \theta \) is \( n\pi \pm \alpha \), where \( n \in I \).
Here, \( \alpha = \frac{\pi}{3} \).
\( \implies \theta = n\pi \pm \frac{\pi}{3} \), where \( n \in I \). This formula helps find all possible angles that satisfy the equation.
In simple words: We start with the given equation and change the right side to a squared sine value. Then, we use a general formula for trigonometric equations to find all the possible angles for \( \theta \).

🎯 Exam Tip: Remember the general solutions for trigonometric equations, especially \( \sin^2 \theta = \sin^2 \alpha \), \( \cos^2 \theta = \cos^2 \alpha \), and \( \tan^2 \theta = \tan^2 \alpha \), which all have the form \( n\pi \pm \alpha \).

Question 2. Find the general solution of \( \sin 2\theta = \cos \theta \).
Answer:
We are given the equation: \( \sin 2\theta = \cos \theta \)
We know that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Substitute this into the equation.
\( \implies 2 \sin \theta \cos \theta = \cos \theta \)
Move all terms to one side to set the equation to zero.
\( \implies 2 \sin \theta \cos \theta - \cos \theta = 0 \)
Factor out \( \cos \theta \).
\( \implies \cos \theta (2 \sin \theta - 1) = 0 \)
This means either \( \cos \theta = 0 \) or \( 2 \sin \theta - 1 = 0 \).
**Case 1: \( \cos \theta = 0 \)**
The general solution for \( \cos \theta = 0 \) is \( \theta = n\pi + \frac{\pi}{2} \), where \( n \in I \). This covers all angles where cosine is zero, like 90°, 270°, etc.
**Case 2: \( 2 \sin \theta - 1 = 0 \)**
\( \implies 2 \sin \theta = 1 \)
\( \implies \sin \theta = \frac{1}{2} \)
We know that \( \sin \frac{\pi}{6} = \frac{1}{2} \).
So, \( \sin \theta = \sin \frac{\pi}{6} \)
The general solution for \( \sin \theta = \sin \alpha \) is \( \theta = n\pi + (-1)^n \alpha \), where \( n \in I \).
\( \implies \theta = n\pi + (-1)^n \frac{\pi}{6} \), where \( n \in I \).
Therefore, the general solution for the given equation is \( \theta = n\pi + \frac{\pi}{2} \) or \( \theta = n\pi + (-1)^n \frac{\pi}{6} \), where \( n \in I \).
In simple words: We change \( \sin 2\theta \) to \( 2 \sin \theta \cos \theta \), then factor the equation. This gives us two simpler equations to solve, one for \( \cos \theta = 0 \) and one for \( \sin \theta = \frac{1}{2} \). We then use standard formulas to find all possible values of \( \theta \) for both cases.

🎯 Exam Tip: When solving equations with different trigonometric functions, try to bring all terms to one side and factorize. This often leads to simpler equations that can be solved using general solution formulas.

Question 3. Solve: \( \sin 5x – \sin 3x – \sin x = 0 \), \( 0^\circ < x < 360^\circ \).
Answer:
We are given the equation: \( \sin 5x - \sin 3x - \sin x = 0 \), for \( 0^\circ < x < 360^\circ \).
First, apply the sum-to-product formula for \( \sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \) to the first two terms.
So, \( \sin 5x - \sin 3x = 2 \cos \left(\frac{5x+3x}{2}\right) \sin \left(\frac{5x-3x}{2}\right) \)
\( \implies 2 \cos \left(\frac{8x}{2}\right) \sin \left(\frac{2x}{2}\right) - \sin x = 0 \)
\( \implies 2 \cos 4x \sin x - \sin x = 0 \)
Now, factor out \( \sin x \) from both terms.
\( \implies \sin x (2 \cos 4x - 1) = 0 \)
This means either \( \sin x = 0 \) or \( 2 \cos 4x - 1 = 0 \).
**Case I: \( \sin x = 0 \)**
The general solution for \( \sin x = 0 \) is \( x = n\pi \), where \( n \in I \).
For the interval \( 0^\circ < x < 360^\circ \) (which is \( 0 < x < 2\pi \) in radians):
If \( n=0 \), \( x = 0 \). But \( x \) must be greater than \( 0 \).
If \( n=1 \), \( x = \pi \). This is within the given range.
If \( n=2 \), \( x = 2\pi \). But \( x \) must be less than \( 2\pi \).
So, from Case I, \( x = \pi \).
**Case II: \( 2 \cos 4x - 1 = 0 \)**
\( \implies 2 \cos 4x = 1 \)
\( \implies \cos 4x = \frac{1}{2} \)
We know that \( \cos \frac{\pi}{3} = \frac{1}{2} \).
So, \( \cos 4x = \cos \frac{\pi}{3} \)
The general solution for \( \cos \theta = \cos \alpha \) is \( \theta = 2n\pi \pm \alpha \), where \( n \in I \).
So, \( 4x = 2n\pi \pm \frac{\pi}{3} \)
Now, divide by 4 to find \( x \).
\( \implies x = \frac{2n\pi}{4} \pm \frac{\pi}{3 \times 4} \)
\( \implies x = \frac{n\pi}{2} \pm \frac{\pi}{12} \), where \( n \in I \).
Now we find the values of \( x \) in the range \( 0 < x < 2\pi \).
For \( n=0 \): \( x = 0 \pm \frac{\pi}{12} \implies x = \frac{\pi}{12} \) ( \( -\frac{\pi}{12} \) is out of range).
For \( n=1 \): \( x = \frac{\pi}{2} \pm \frac{\pi}{12} \)
\( x = \frac{\pi}{2} + \frac{\pi}{12} = \frac{6\pi + \pi}{12} = \frac{7\pi}{12} \)
\( x = \frac{\pi}{2} - \frac{\pi}{12} = \frac{6\pi - \pi}{12} = \frac{5\pi}{12} \)
For \( n=2 \): \( x = \pi \pm \frac{\pi}{12} \)
\( x = \pi + \frac{\pi}{12} = \frac{12\pi + \pi}{12} = \frac{13\pi}{12} \)
\( x = \pi - \frac{\pi}{12} = \frac{12\pi - \pi}{12} = \frac{11\pi}{12} \)
For \( n=3 \): \( x = \frac{3\pi}{2} \pm \frac{\pi}{12} \)
\( x = \frac{3\pi}{2} + \frac{\pi}{12} = \frac{18\pi + \pi}{12} = \frac{19\pi}{12} \)
\( x = \frac{3\pi}{2} - \frac{\pi}{12} = \frac{18\pi - \pi}{12} = \frac{17\pi}{12} \)
For \( n=4 \): \( x = 2\pi \pm \frac{\pi}{12} \). This gives values outside the range \( 0 < x < 2\pi \).
Combining all values, the solutions for \( x \) in the given range are: \( \pi, \frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12} \). These are all the angles within the 0 to 360-degree range that satisfy the equation.
In simple words: First, we use a math rule to simplify the equation, turning two sine terms into one. Then, we factor the equation to get two simpler parts. We solve each part separately to find all angles between 0 and 360 degrees that make the equation true.

🎯 Exam Tip: Always remember to check your solutions against the given domain (e.g., \( 0^\circ < x < 360^\circ \)) and discard any values that fall outside it. Also, be careful with the general solution formulas for \( \sin x = 0 \) and \( \cos x = \cos \alpha \).

Question 4. Find the general solution of \( \tan^2 \theta = \frac { 1 }{ 3 } \), and hence find those values of \( \theta \) for which \( - \pi \leq \theta \leq \pi \).
Answer:
We are given the equation: \( \tan^2 \theta = \frac{1}{3} \)
We know that \( \frac{1}{3} \) can be written as \( \left(\frac{1}{\sqrt{3}}\right)^2 \).
Also, \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \).
So, \( \tan^2 \theta = \left(\tan \frac{\pi}{6}\right)^2 = \tan^2 \frac{\pi}{6} \).
When \( \tan^2 \theta = \tan^2 \alpha \), the general solution is \( \theta = n\pi \pm \alpha \), where \( n \in I \).
\( \implies \theta = n\pi \pm \frac{\pi}{6} \), where \( n \in I \). This is the general solution for all possible values of \( \theta \).

Now, we need to find the specific values of \( \theta \) in the interval \( - \pi \leq \theta \leq \pi \).
**For \( n=0 \):**
\( \theta = 0\pi \pm \frac{\pi}{6} \implies \theta = \pm \frac{\pi}{6} \). Both \( \frac{\pi}{6} \) and \( -\frac{\pi}{6} \) are within the range \( [-\pi, \pi] \).
**For \( n=1 \):**
\( \theta = 1\pi \pm \frac{\pi}{6} \)
\( \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \). This value is greater than \( \pi \), so it is outside the range.
\( \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \). This value is within the range \( [-\pi, \pi] \).
**For \( n=-1 \):**
\( \theta = -1\pi \pm \frac{\pi}{6} \)
\( \theta = -\pi + \frac{\pi}{6} = \frac{-6\pi + \pi}{6} = -\frac{5\pi}{6} \). This value is within the range \( [-\pi, \pi] \).
\( \theta = -\pi - \frac{\pi}{6} = \frac{-6\pi - \pi}{6} = -\frac{7\pi}{6} \). This value is less than \( -\pi \), so it is outside the range.
Therefore, the values of \( \theta \) for which \( -\pi \leq \theta \leq \pi \) are \( \pm \frac{\pi}{6}, \pm \frac{5\pi}{6} \). These specific angles fulfill the condition for the given interval.
In simple words: First, we find the general solution for the given tangent equation by recognizing \( \frac{1}{3} \) as \( (\tan \frac{\pi}{6})^2 \). Then, we substitute different integer values for \( n \) into the general solution to find all angles that fall between \( -\pi \) and \( \pi \).

🎯 Exam Tip: Always specify the domain for which you are finding the solutions. When dealing with \( n\pi \pm \alpha \), systematically test integer values of \( n \) (0, 1, -1, 2, -2, etc.) until the generated values fall outside the given range.

Question 6. Solve the equation \( \sqrt{3} \cos x + \sin x = 1 \) for \( -2\pi < x < 2\pi \).
Answer:
We are given the equation: \( \sqrt{3} \cos x + \sin x = 1 \), for \( -2\pi < x < 2\pi \).
To solve this, we can divide the entire equation by \( \sqrt{a^2+b^2} \), where \( a=\sqrt{3} \) and \( b=1 \).
\( \sqrt{a^2+b^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2 \).
Divide the equation by 2:
\( \implies \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x = \frac{1}{2} \)
We know that \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \) and \( \sin \frac{\pi}{6} = \frac{1}{2} \).
Substitute these values into the equation:
\( \implies \cos \frac{\pi}{6} \cos x + \sin \frac{\pi}{6} \sin x = \frac{1}{2} \)
This matches the cosine addition formula: \( \cos(A-B) = \cos A \cos B + \sin A \sin B \).
So, \( \cos \left(x - \frac{\pi}{6}\right) = \frac{1}{2} \)
We know that \( \cos \frac{\pi}{3} = \frac{1}{2} \).
\( \implies \cos \left(x - \frac{\pi}{6}\right) = \cos \frac{\pi}{3} \)
The general solution for \( \cos \theta = \cos \alpha \) is \( \theta = 2n\pi \pm \alpha \), where \( n \in I \).
Here, \( \theta = x - \frac{\pi}{6} \) and \( \alpha = \frac{\pi}{3} \).
\( \implies x - \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{3} \)
Now, solve for \( x \):
\( \implies x = 2n\pi + \frac{\pi}{6} \pm \frac{\pi}{3} \), where \( n \in I \). This formula gives all possible solutions.
We have two cases for \( x \):
**Case A:** \( x = 2n\pi + \frac{\pi}{6} + \frac{\pi}{3} \)
\( x = 2n\pi + \frac{\pi+2\pi}{6} = 2n\pi + \frac{3\pi}{6} = 2n\pi + \frac{\pi}{2} \)
**Case B:** \( x = 2n\pi + \frac{\pi}{6} - \frac{\pi}{3} \)
\( x = 2n\pi + \frac{\pi-2\pi}{6} = 2n\pi - \frac{\pi}{6} \)

Now, we find values for \( x \) in the range \( -2\pi < x < 2\pi \).
**Using Case A: \( x = 2n\pi + \frac{\pi}{2} \)**
If \( n=0 \), \( x = 0\pi + \frac{\pi}{2} = \frac{\pi}{2} \). This is in the range.
If \( n=1 \), \( x = 2\pi + \frac{\pi}{2} = \frac{5\pi}{2} \). This is outside the range.
If \( n=-1 \), \( x = -2\pi + \frac{\pi}{2} = -\frac{3\pi}{2} \). This is in the range.

**Using Case B: \( x = 2n\pi - \frac{\pi}{6} \)**
If \( n=0 \), \( x = 0\pi - \frac{\pi}{6} = -\frac{\pi}{6} \). This is in the range.
If \( n=1 \), \( x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \). This is in the range.
If \( n=-1 \), \( x = -2\pi - \frac{\pi}{6} = -\frac{13\pi}{6} \). This is outside the range.

Combining all values, the solutions for \( x \) in the given range \( -2\pi < x < 2\pi \) are: \( \frac{\pi}{2}, -\frac{3\pi}{2}, -\frac{\pi}{6}, \frac{11\pi}{6} \). These are all the specific angles that solve the equation within the defined limits.
In simple words: We first rewrite the equation using a trick where we divide by a specific number (2 in this case) to make it look like a cosine addition formula. This simplifies the equation, allowing us to find the general solutions for \( x \). Finally, we pick out only those solutions that fit within the given range of \( -2\pi \) to \( 2\pi \).

🎯 Exam Tip: For equations of the form \( a \cos x + b \sin x = c \), always divide by \( \sqrt{a^2+b^2} \) to transform it into a single trigonometric function, making it easier to solve. Remember to check both positive and negative \( n \) values when finding solutions within a specific interval.