GSEB Class 11 Maths Solutions Chapter 3 Trigonometric Functions Exercise 3.4

Find the principal and general solutions of the following equations:

Question 1. tan x = \( \sqrt{3} \)
Answer: We have \( \tan x = \sqrt{3} \). We know that \( \tan 60^\circ = \sqrt{3} \). Therefore, the principal value of \( x \) is \( 60^\circ \), which is also written as \( \frac{\pi}{3} \) radians. If we have \( \tan x = \tan \alpha \), where \( \alpha \) is the principal value, then the general solution for \( x \) is given by \( n\pi + \alpha \). So, the general value of \( x \) becomes \( n\pi + \frac{\pi}{3} \).
In simple words: The main angle where tan x is \( \sqrt{3} \) is \( 60^\circ \) or \( \frac{\pi}{3} \) radians. The full set of solutions is \( n\pi \) plus this main angle.

Exam Tip: Remember to convert degree values to radians for general solutions in trigonometry, as it is standard practice. Always identify the principal value first to find the general solution.

Question 2. sec x = 2
Answer: We have \( \sec x = 2 \). This implies that \( \cos x = \frac{1}{2} \). We know that \( \cos 60^\circ = \frac{1}{2} \). So, the principal value is \( 60^\circ \), which is \( \frac{\pi}{3} \) radians. For an equation like \( \cos \theta = \cos \alpha \), the general solution is \( \theta = 2n\pi \pm \alpha \). Thus, the general value of \( x \) is \( 2n\pi \pm \frac{\pi}{3} \).
In simple words: When sec x is 2, cos x is \( \frac{1}{2} \). The main angle for this is \( 60^\circ \). The complete set of angles is \( 2n\pi \) plus or minus this angle.

Exam Tip: When working with secant, cosecant, or cotangent, always convert the equation to cosine, sine, or tangent, respectively, for easier calculation of principal and general solutions.

Question 3. cot x = \( -\sqrt{3} \)
Answer: We have \( \cot x = -\sqrt{3} \). This implies \( \tan x = -\frac{1}{\sqrt{3}} \). We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \). To get a negative value, we use \( \tan (180^\circ - \theta) = -\tan \theta \). So, \( \tan (180^\circ - 30^\circ) = -\tan 30^\circ = -\frac{1}{\sqrt{3}} \). This means \( \tan 150^\circ = -\frac{1}{\sqrt{3}} \). The principal value of \( x \) is \( 150^\circ \), which is \( \frac{5\pi}{6} \) radians. The general solution for \( \tan x = \tan \alpha \) is \( x = n\pi + \alpha \). Therefore, the general value of \( x \) is \( n\pi + \frac{5\pi}{6} \).
In simple words: If cot x is \( -\sqrt{3} \), then tan x is \( -\frac{1}{\sqrt{3}} \). The main angle that gives this is \( 150^\circ \). To find all solutions, add \( n\pi \) to this main angle.

Exam Tip: For negative values of trigonometric functions, determine the quadrant where the function is negative and find the smallest positive angle in that quadrant to use as the principal value.

Question 4. cosec x = - 2
Answer: We have \( \operatorname{cosec} x = -2 \). This means \( \sin x = -\frac{1}{2} \). We know that \( \sin 30^\circ = \frac{1}{2} \), so \( \sin (-30^\circ) = -\frac{1}{2} \). The principal value of \( x \) is \( -30^\circ \), which is \( -\frac{\pi}{6} \) radians. For an equation \( \sin x = \sin \alpha \), the general solution is \( x = n\pi + (-1)^n \alpha \). So, the general value of \( x \) is \( n\pi + (-1)^n \left(-\frac{\pi}{6}\right) \). This can also be written as \( n\pi - (-1)^n \frac{\pi}{6} \).
In simple words: When cosec x is -2, sin x is \( -\frac{1}{2} \). The principal angle is \( -30^\circ \) or \( -\frac{\pi}{6} \) radians. The full range of solutions includes \( n\pi \) and alternates the sign of the principal angle based on \( n \).

Exam Tip: When the principal value is negative, include the negative sign carefully in the general solution formula \( x = n\pi + (-1)^n \alpha \). Alternatively, you can use the positive coterminal angle for the principal value (e.g., \( 330^\circ \) or \( \frac{11\pi}{6} \)).

Find the solution for each of the following equations:

Question 5. cos 4x = cos 2x
Answer: We are given the equation \( \cos 4x = \cos 2x \). We can rewrite this as \( \cos 4x - \cos 2x = 0 \). Using the sum-to-product formula \( \cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \), we get:
\( -2 \sin \left(\frac{4x+2x}{2}\right) \sin \left(\frac{4x-2x}{2}\right) = 0 \)
\( -2 \sin (3x) \sin (x) = 0 \)
\( \implies \sin (3x) \sin (x) = 0 \) This means either \( \sin 3x = 0 \) or \( \sin x = 0 \).
If \( \sin 3x = 0 \), then \( 3x = n\pi \implies x = \frac{n\pi}{3} \).
If \( \sin x = 0 \), then \( x = n\pi \). These are the general solutions for the given equation.
In simple words: When \( \cos 4x \) equals \( \cos 2x \), it means their difference is zero. We use a formula to change this into a product of sines. This gives two possibilities: either \( \sin 3x \) is zero, or \( \sin x \) is zero, leading to the solutions \( x = \frac{n\pi}{3} \) or \( x = n\pi \).

Exam Tip: For equations of the form \( \cos A = \cos B \), you can directly use the general solution \( A = 2n\pi \pm B \). Alternatively, transforming into a product of sines using sum-to-product identities is also a valid approach.

Question 6. cos 3x + cos x - cos 2x = 0
Answer: We are given the equation \( \cos 3x + \cos x - \cos 2x = 0 \). First, apply the sum-to-product formula \( \cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) \) to the first two terms:
\( 2 \cos \left(\frac{3x+x}{2}\right) \cos \left(\frac{3x-x}{2}\right) - \cos 2x = 0 \)
\( 2 \cos (2x) \cos (x) - \cos 2x = 0 \) Now, factor out \( \cos 2x \):
\( \cos 2x (2 \cos x - 1) = 0 \) This means either \( \cos 2x = 0 \) or \( 2 \cos x - 1 = 0 \).
Case 1: If \( \cos 2x = 0 \), then \( 2x = (2n+1)\frac{\pi}{2} \implies x = (2n+1)\frac{\pi}{4} \).
Case 2: If \( 2 \cos x - 1 = 0 \), then \( \cos x = \frac{1}{2} \). We know \( \cos \frac{\pi}{3} = \frac{1}{2} \). So, for \( \cos x = \cos \frac{\pi}{3} \), the general solution is \( x = 2n\pi \pm \frac{\pi}{3} \). These are the general solutions for the given equation.
In simple words: We start by combining \( \cos 3x \) and \( \cos x \) using a formula, then we take out the common term \( \cos 2x \). This gives two separate equations to solve: one where \( \cos 2x \) is zero, and another where \( \cos x \) is \( \frac{1}{2} \). Each gives a set of general solutions for \( x \).

Exam Tip: Look for opportunities to factor common trigonometric terms after applying sum-to-product or product-to-sum identities. This often simplifies the equation into easier-to-solve components.

Question 7. sin 2x + cos x = 0
Answer: We have the equation \( \sin 2x + \cos x = 0 \). We know that \( \sin 2x = 2 \sin x \cos x \). Substitute this into the equation:
\( 2 \sin x \cos x + \cos x = 0 \) Factor out the common term \( \cos x \):
\( \cos x (2 \sin x + 1) = 0 \) This implies either \( \cos x = 0 \) or \( 2 \sin x + 1 = 0 \).
Case 1: If \( \cos x = 0 \), then \( x = (2n+1)\frac{\pi}{2} \).
Case 2: If \( 2 \sin x + 1 = 0 \), then \( \sin x = -\frac{1}{2} \). We know \( \sin \frac{\pi}{6} = \frac{1}{2} \). Since \( \sin x \) is negative, \( x \) must be in the third or fourth quadrant. The principal value is \( -\frac{\pi}{6} \) (or \( \frac{7\pi}{6} \)). For \( \sin x = \sin \alpha \), the general solution is \( x = n\pi + (-1)^n \alpha \). So, \( x = n\pi + (-1)^n \left(-\frac{\pi}{6}\right) \). This can be expressed as \( x = n\pi - (-1)^n \frac{\pi}{6} \).
In simple words: First, change \( \sin 2x \) to \( 2 \sin x \cos x \). Then, pull out \( \cos x \) as a common factor. This gives two separate equations: \( \cos x = 0 \) and \( \sin x = -\frac{1}{2} \). Solve each to get the overall solutions for \( x \).

Exam Tip: Always use double angle identities like \( \sin 2x = 2 \sin x \cos x \) to simplify equations and look for common factors. Be careful with the sign of the principal angle when dealing with negative trigonometric values.

Question 8. sec\(^2\) 2x = 1 - tan 2x
Answer: We are given the equation \( \sec^2 2x = 1 - \tan 2x \). We know the identity \( \sec^2 \theta = 1 + \tan^2 \theta \). Apply this to the left side:
\( 1 + \tan^2 2x = 1 - \tan 2x \) Rearrange the terms to form a quadratic equation in \( \tan 2x \):
\( \tan^2 2x + \tan 2x = 0 \) Factor out \( \tan 2x \):
\( \tan 2x (\tan 2x + 1) = 0 \) This means either \( \tan 2x = 0 \) or \( \tan 2x + 1 = 0 \).
Case 1: If \( \tan 2x = 0 \), then \( 2x = n\pi \implies x = \frac{n\pi}{2} \).
Case 2: If \( \tan 2x + 1 = 0 \), then \( \tan 2x = -1 \). We know \( \tan \frac{\pi}{4} = 1 \). Since \( \tan 2x \) is negative, \( 2x \) is in the second or fourth quadrant. The principal value is \( \frac{3\pi}{4} \), because \( \tan (\pi - \frac{\pi}{4}) = -\tan \frac{\pi}{4} = -1 \). So, for \( \tan 2x = \tan \frac{3\pi}{4} \), the general solution is \( 2x = n\pi + \frac{3\pi}{4} \).
\( \implies x = \frac{n\pi}{2} + \frac{3\pi}{8} \). These are the general solutions for the given equation.
In simple words: Replace \( \sec^2 2x \) with \( 1 + \tan^2 2x \). Then, move all terms to one side to get a simple quadratic equation in terms of \( \tan 2x \). Factor it to find two possibilities: \( \tan 2x = 0 \) or \( \tan 2x = -1 \). Solve each for \( 2x \), then divide by 2 to get \( x \).

Exam Tip: Always look for fundamental trigonometric identities (like \( \sec^2 \theta = 1 + \tan^2 \theta \)) that can simplify the equation and allow for algebraic manipulation, such as factoring.

Question 9. sin x + sin 3x + sin 5x = 0
Answer: We have the equation \( \sin x + \sin 3x + \sin 5x = 0 \). Rearrange the terms and group \( \sin 5x \) and \( \sin x \) together:
\( (\sin 5x + \sin x) + \sin 3x = 0 \) Apply the sum-to-product formula \( \sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) \) to the grouped terms:
\( 2 \sin \left(\frac{5x+x}{2}\right) \cos \left(\frac{5x-x}{2}\right) + \sin 3x = 0 \)
\( 2 \sin (3x) \cos (2x) + \sin 3x = 0 \) Factor out the common term \( \sin 3x \):
\( \sin 3x (2 \cos 2x + 1) = 0 \) This means either \( \sin 3x = 0 \) or \( 2 \cos 2x + 1 = 0 \).
Case 1: If \( \sin 3x = 0 \), then \( 3x = n\pi \implies x = \frac{n\pi}{3} \).
Case 2: If \( 2 \cos 2x + 1 = 0 \), then \( \cos 2x = -\frac{1}{2} \). We know \( \cos \frac{\pi}{3} = \frac{1}{2} \). Since \( \cos 2x \) is negative, \( 2x \) is in the second or third quadrant. The principal value is \( \frac{2\pi}{3} \), because \( \cos (\pi - \frac{\pi}{3}) = -\cos \frac{\pi}{3} = -\frac{1}{2} \). So, for \( \cos 2x = \cos \frac{2\pi}{3} \), the general solution is \( 2x = 2n\pi \pm \frac{2\pi}{3} \).
Dividing by 2 gives: \( x = n\pi \pm \frac{\pi}{3} \). These are the general solutions for the given equation.
In simple words: First, group \( \sin x \) and \( \sin 5x \) and use a sum-to-product formula to simplify them. You will then find a common factor of \( \sin 3x \). Factor this out, leading to two separate equations: \( \sin 3x = 0 \) and \( 2 \cos 2x + 1 = 0 \). Solve each one to get the final general solutions.

Exam Tip: When dealing with three or more trigonometric terms in an equation, try grouping two terms that simplify well using sum-to-product identities, often aiming for a common factor with the remaining term.