# GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4

Find the principal and general solutions of the following equations:
1. tan x = $$\sqrt{3}$$
2. sec x = 2
3. cot x = – $$\sqrt{3}$$
4. cosec x = – 2
Solutions to questions 1 – 4:
1. tan x = $$\sqrt{3}$$ = tan 60°.
∴ Principal value of x = 60° = $$\frac{π}{3}$$ radians.
If tan x = tan α, when a is the principal value of θ,
then x = nπ + α.
∴ General value of x = nπ + $$\frac{π}{3}$$

2. sec x = 2 = sec 60° or cos x = $$\frac{1}{2}$$ = cos 60°.
∴ Principal value = 60°= $$\frac{π}{3}$$ radians.
For cos θ = cos α, θ = 2nπ ± α.
∴ General value of x = 2nn ± $$\frac{π}{3}$$.

3. cot x = – $$\sqrt{3}$$ ⇒ tan x = – $$\frac{1}{\sqrt{3}}$$
Now, tan 30° = $$\frac{1}{\sqrt{3}}$$ ⇒ tan (180° – 30°) = – tan 30°.
= – $$\frac{1}{\sqrt{3}}$$ or tan 150° = – $$\frac{1}{\sqrt{3}}$$.
Thus, principle value of x = 150° = $$\frac{5π}{6}$$ radians.
∴ General value of x = nπ + α
= nπ + $$\frac{5π}{6}$$

4. cosec x = – 2 or sin x = – $$\frac{1}{2}$$
sin 30° = $$\frac{1}{2}$$ or sin (- 30°) = – sin 30° = – $$\frac{1}{2}$$.
∴ Principal value of x = – 30° = – $$\frac{π}{6}$$.
So, general value of x = nπ + (- 1)nα
= nπ + (- 1)n (- $$\frac{π}{6}$$) = nπ – (- 1)n ($$\frac{π}{6}$$).

Find the solution for each of the following equations:
5. cos 4x = cos 2x
6. cos 3x + cos x – cos 2x = 0
7. sin 2x + cos x = 0
8. sec2 2x = 1 – tan 2x
9. sin x + sin 3x + sin 5x = 0
Solutions to questions 5 – 10:

5. cos 4x = cos 2x
or cos 2x – cos 4x = 0.
or 2 sin $$\frac{2x+4x}{2}$$sin $$\frac{4x-2x}{2}$$ = 0
or 2sin 3x sin x = 0.
If sin 3x = 0, then 3x = nπ or x = $$\frac{nπ}{3}$$.
If sin x = 0, then x = nπ.

6. cos 3x + cos x – cos 2x = 0
or 2cos $$\frac{3x+x}{2}$$cos $$\frac{3x-x}{2}$$ – cos 2x = 0.
or 2 cos 2x cosx – cos 2x = 0
or cos 2x(2 cos x – 1) = 0.
If cos 2x = 0, then 2x = (2n + 1)$$\frac{π}{2}$$ ⇒ x = (2n + 1)$$\frac{π}{4}$$.
If 2cos x – 1 = 0, cos x = $$\frac{1}{2}$$ = cos 60° = cos$$\frac{π}{3}$$.
⇒ x = 2nπ ± $$\frac{π}{3}$$.

7. sin 2x + cos x = 0
or 2sin x cos x + cos x = 0
or cos x(2sin x + 1) = 0.
sin x = – $$\frac{1}{2}$$ = sin(π + $$\frac{π}{6}$$) = sin $$\frac{7π}{6}$$ ⇒ x = nπ + (- 1)n $$\frac{7π}{6}$$.

8. sec2 2x = 1 – tan 2x
⇒ 1 + tan2 2x = 1 – tan 2x = 0
⇒ tan2 2x + tan 2x = 0
⇒ tan 2x(tan 2x + 1) = 0.
If tan 2x = 0, then 2x = nπ or x = $$\frac{nπ}{2}$$
If tan 2x + 1 = 0, then tan 2x = – 1 = tan (π – $$\frac{π}{4}$$) = tan $$\frac{3π}{4}$$
⇒ 2x = nπ + $$\frac{3π}{4}$$ or x = $$\frac{nπ}{2}$$ + $$\frac{3π}{8}$$.

9. sin x + sin 3x + sin 5x = 0
or (sin 5x + sin x) + sin 3x = 0.
or 2 sin $$\frac{5x+x}{2}$$cos $$\frac{5x-x}{2}$$ + sin 3x = 0
or 2 sin 3x cos 2x + sin 3x = 0
or sin 3x(2 cos 2x + 1) = 0.
If sin 3x = 0, then 3x = nπ or x = $$\frac{nπ}{3}$$
If 2cos 2x + 1 = 0, then cos 2x = – $$\frac{1}{2}$$ = cos (π – $$\frac{π}{3}$$) = cos $$\frac{2π}{3}$$
∴ 2x = 2nπ ± $$\frac{2π}{3}$$ or x = nπ ± $$\frac{π}{3}$$.