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 ± \(\frac{π}{3}\).