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}).