ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

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ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

Question 1.
Find the principal solutions of the following equations :
(i) sin x = \(\frac{\sqrt{3}}{2}\)
(ii) cos x = 1
(iii) tan x = \(\frac{1}{\sqrt{3}}\)
(iv) tan x = – \(\frac{1}{\sqrt{3}}\)
(v) cos x = – 1
(vi) cosec x = – 1
(vii) sec x = – √2
(viii) tan x = – 1
(ix) cosec x = – 2.
Solution:
(i) We know that, the solutions lying between 0 to 2π (0 ≤ x < 2π) are principal solutions.
Given sin x = \(\frac{\sqrt{3}}{2}\)
we know that,
sin \(\frac{\pi}{3}\) = \(\frac{\sqrt{3}}{2}\)
and sin (π – \(\frac{\pi}{2}\)) = sin \(\frac{\pi}{3}\) = \(\frac{\sqrt{3}}{2}\)
∴ sin \(\frac{\pi}{3}\) = \(\frac{\sqrt{3}}{2}\)
and sin \(\frac{2 \pi}{3}\) = \(\frac{\sqrt{3}}{2}\)
thus the principal solutions are \(\frac{\pi}{3}\), \(\frac{2 \pi}{3}\).

(ii) Given cos x = 1
since cos 0 = 1
and cos 2π = 1
but 2π ∉ [0, 2π)
∴ The principal soln. be 0.

(iii) Given tan x = \(\frac{1}{\sqrt{3}}\)
We know that,
tan \(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)
and tan (π + \(\frac{\pi}{6}\)) = tan \(\frac{\pi}{6}\)
= \(\frac{1}{\sqrt{3}}\)
∴ tan \(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)
and tan (\(\frac{7 \pi}{6}\)) = \(\frac{1}{\sqrt{3}}\)
Thus, the principal solutions are \(\frac{\pi}{6}\) and \(\frac{7 \pi}{6}\).

(iv) Given tan x = – \(\frac{1}{\sqrt{3}}\)
We know that,
tan \(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)
∴ tan (π – \(\frac{\pi}{6}\)) = – tan \(\frac{\pi}{6}\)
= – \(\frac{1}{\sqrt{3}}\)
and tan (2π – \(\frac{\pi}{6}\)) = – tan \(\frac{\pi}{6}\)
= – \(\frac{1}{\sqrt{3}}\)
Thus tan \(\frac{5 \pi}{6}\) = – \(\frac{1}{\sqrt{3}}\)
and tan \(\left(\frac{11 \pi}{6}\right)=-\frac{1}{\sqrt{3}}\)
Hence, \(\frac{5 \pi}{6}\) and \(\frac{11 \pi}{6}\) are the two principal solutions.

(v) Given cos x = – 1
We know that,
cos 0 = 1
∴ cos (π – 0) = – cos 0 = – 1
Thus, the only principal soln. be π.

(vi) Given cosec x = – 1
⇒ sin x = – 1
We know that,
sin \(\frac{\pi}{2}\) = 1
∴ sin (π + \(\frac{\pi}{2}\)) = – sin \(\frac{\pi}{2}\) = – 1
or sin (2π – \(\frac{\pi}{2}\)) = – sin \(\frac{\pi}{2}\) = – 1
Thus, sin \(\frac{3 \pi}{2}\) = – 1
Hence \(\frac{3 \pi}{2}\) be the only principal soln.

(vii) Given sec x = – √2
⇒ cos x = – \(\frac{1}{\sqrt{2}}\)
We know that,
cos \(\frac{\pi}{4}=\frac{1}{\sqrt{2}}\)
∴ cos (π – \(\frac{\pi}{4}\)) = – cos \(\frac{\pi}{4}\)
= – \(\frac{1}{\sqrt{2}}\)
and cos (π + \(\frac{\pi}{4}\)) = – cos \(\frac{\pi}{4}\)
= – \(\frac{1}{\sqrt{2}}\)
Thus, cos \(\frac{3 \pi}{4}=-\frac{1}{\sqrt{2}}\)
and cos \(\frac{5 \pi}{4}=-\frac{1}{\sqrt{2}}\)
Hence, \(\frac{3 \pi}{4}\) and \(\frac{5 \pi}{4}\) are the principal solutions.

(viii) Given tan x = – 1 …( 1)
we know that,
tan \(\frac{\pi}{4}\) = 1
tan (π – \(\frac{\pi}{4}\)) = – tan \(\frac{\pi}{4}\) = – 1
and tan (2π – \(\frac{\pi}{4}\)) = – tan \(\frac{\pi}{4}\) = – 1
Thus, tan \(\frac{3 \pi}{4}\) = – 1
and tan (\(\frac{7 \pi}{4}\)) = – 1
Hence the principal solutions of eqn. (1) are \(\frac{3 \pi}{4}\) and \(\frac{7 \pi}{4}\).

(ix) Given √3 cosec x = – 2
⇒ cosec x = – \(\frac{2}{\sqrt{3}}\)
We know that
sin x = – \(\frac{\sqrt{3}}{2}\)
We know that,
sin \(\frac{\pi}{3}\) = \(\frac{\sqrt{3}}{2}\)
sin (π + \(\frac{\pi}{3}\)) = – sin \(\frac{\pi}{3}\)
= – \(\frac{\sqrt{3}}{2}\)
and sin (2π – \(\frac{\pi}{3}\)) = – sin \(\frac{\pi}{3}\)
= – \(\frac{\sqrt{3}}{2}\)
Thus sin \(\frac{4 \pi}{3}\) = – \(\frac{\sqrt{3}}{2}\)
and sin \(\frac{5 \pi}{3}=-\frac{\sqrt{3}}{2}\)
Hence the required solution of given eqn. are \(\frac{4 \pi}{3}\) and \(\frac{5 \pi}{3}\).

Find the general solutions of the following (2 to 18) equations :

Question 2.
(i) sin 3x = 0
(ii) sin \(\frac{5 x}{2}\) = 0
(iii) cos (x + \(\frac{\pi}{12}\)) = 0
(iv) cos (x + \(\frac{\pi}{4}\)) = 0
(v) tan \(\frac{3 x}{4}\) = 0
(vi) tau (x – \(\frac{x}{4}\)) = 0
Solution:
(i) Given sin 3x = 0 = sin 0
⇒ 3x = n π ∀ n ∈ I
x = \(\frac{n \pi}{4}\) ∀ n ∈ I

(ii) Given sin \(\frac{5 x}{2}\) = 0
⇒ \(\frac{5 x}{2}\) = n π, n ∈ I
⇒ x = \(\frac{2 n \pi}{5}\) ∀ n ∈ I

(iii) Given sin (x + \(\frac{\pi}{12}\)) = 0
⇒ x + \(\frac{\pi}{12}\) = nπ ∀ n ∈ I
[∵ sin θ = 0
⇒ θ = nπ ∀ n ∈ I]
⇒ x = nπ – \(\frac{\pi}{12}\), n ∈ I

(iv) cos (x + \(\frac{\pi}{4}\)) = 0
x + \(\frac{\pi}{4}\) = (2n + 1) \(\frac{\pi}{2}\), n ∈ I
⇒ x = (2n + 1) \(\frac{\pi}{2}\) – \(\frac{\pi}{4}\), n ∈ I
⇒ x = nπ + \(\frac{\pi}{4}\), n ∈ I

(v) Given tan \(\frac{3x}{4}\) = 0
⇒ \(\frac{3x}{4}\) = nπ, n ∈ I
⇒ x = \(\frac{4 n \pi}{3}\), n ∈ I

(vi) Given tan (x – \(\frac{\pi}{4}\)) = 0
⇒ x – \(\frac{\pi}{4}\) = nπ ∀ n ∈ I
⇒ x = nπ + \(\frac{\pi}{4}\) ∀ n ∈ I

Question 3.
(i) cos x = \(\frac{\sqrt{3}}{2}\)
(ii) cot x = \(\frac{1}{\sqrt{3}}\)
(iii) sec x = 2
(iv) sin x = – \(\frac{\sqrt{3}}{2}\)
(v) cot x = – √3
(vi) cosec x = – 2.
Solution:
(i) Given cos x = \(\frac{\sqrt{3}}{2}\)
⇒ cos x = cos \(\frac{\pi}{6}\)
⇒ x = 2nπ ± \(\frac{\pi}{6}\) ∀ n ∈ I
[∵ cos θ = cos α
⇒ θ = 2nπ ± α, n ∈ I]
which is the required general soln.

(ii) Given cot x = \(\frac{1}{\sqrt{3}}\)
⇒ tan x = √3 = tan \(\frac{\pi}{3}\)
⇒ x = nπ + \(\frac{\pi}{3}\) ∀ n ∈ I
which is the required soln.

(iii) Given sec x = 2
⇒ cos x = \(\frac{1}{2}\)
= cos \(\frac{\pi}{3}\)
⇒ x = 2nπ ± \(\frac{\pi}{3}\) ∀ n ∈ I

(iv) Given sin x = – \(\frac{\sqrt{3}}{2}\)
= – sin \(\frac{\pi}{3}\)
= sin (π + \(\frac{\pi}{3}\))
⇒ sin x = sin \(\frac{4 \pi}{3}\)
⇒ x = nπ + (- 1)ⁿ \(\frac{4 \pi}{3}\) ∀ n ∈ I
[∵ sin θ = sin α
⇒ θ = nπ + (- 1)ⁿ \(\frac{4 \pi}{3}\) ∀ n ∈ I]

(v) Given cot x = – √3
⇒ tan x = – \(\frac{1}{\sqrt{3}}\)
= – tan \(\frac{\pi}{2}\)
= tan (π – \(\frac{\pi}{6}\))
⇒ tan x = tan \(\frac{5 \pi}{6}\)
⇒ x = nπ + \(\frac{5 \pi}{6}\) ∀ n ∈ I
[∵ tan θ = tan α
⇒ θ = nπ + α ∀ n ∈ I]

(vi) Given cosec x = – 2
⇒ sin x = – \(\frac{1}{2}\)
= – sin \(\frac{\pi}{6}\)
= sin (π + \(\frac{\pi}{6}\))
⇒ sin x = sin \(\frac{7 \pi}{6}\)
⇒ x = nπ + (- 1)ⁿ \(\frac{7 \pi}{6}\) ∀ n ∈ I

Question 4.
(i) √2 sin x + 1 = 0
(ii) 1 + tan x = 0
(iii) √3 sec x + 2 = 0
Solution:
(i) Given √2 sin x + 1 = 0
⇒ sin x = – \(\frac{1}{\sqrt{2}}\)
= – sin \(\frac{\pi}{4}\)
= sin (π + \(\frac{\pi}{4}\))
⇒ sin x = sin \(\frac{5 \pi}{4}\)
⇒ x = nπ + (- 1)ⁿ \(\frac{5 \pi}{4}\) ; n ∈ I
which is the required solution.

(ii) Given 1+ tan x = 0
⇒ tan x = – 1
= – tan \(\frac{\pi}{4}\)
⇒ tan x = – tan (π – \(\frac{\pi}{4}\))
⇒ tan x = tan \(\frac{\pi}{4}\)
⇒ x = nπ + \(\frac{3 \pi}{4}\) where n ∈ I

(iii) Given √3 sec x + 2 = 0
⇒ sec x = – \(\frac{2}{\sqrt{3}}\)
⇒ cos x = – \(\frac{\sqrt{3}}{2}\)
⇒ cos x = – cos \(\frac{\pi}{6}\)
= cos (π – \(\frac{\pi}{6}\))
⇒ cos x = cos \(\frac{5 \pi}{6}\)
⇒ x = 2nπ ± \(\frac{5 \pi}{6}\) ; ∀ n ∈ I

Question 5.
(i) tan² (x + \(\frac{\pi}{3}\)) = 0
(ii) 4 cos² x = 3
(iii) 3 cosec² x = 4.
Solution:
(i) Given tan² (x + \(\frac{\pi}{3}\)) = 0
= tan² 0
⇒ x + \(\frac{\pi}{3}\) = nπ – \(\frac{\pi}{3}\) ∀ n ∈ I
[∵ tan² θ = tan² α
⇒ θ = nπ ± α ; n ∈ I]
⇒ x = nπ – \(\frac{\pi}{3}\) ∀ n ∈ I

(ii) Given 4 cos² x = 3
⇒ cos² x = \(\frac{3}{4}\)
= \(\left(\frac{\sqrt{3}}{2}\right)^2\)
= cos² \(\frac{\pi}{6}\)
⇒ x = nπ ± \(\frac{\pi}{6}\) ; n ∈ I
[∵ cos² θ = cos² α
⇒ θ = nπ ± α ; n ∈ I]

(iii) Given 3 cosec² x = 4
cosec² x = \(\frac{4}{3}\)
sin² x = \(\frac{3}{4}=\left(\frac{\sqrt{3}}{2}\right)^2\)
sin² x = sin² \(\frac{\pi}{3}\)
x = nπ ± \(\frac{\pi}{3}\) ; n ∈ I

Question 6.
(i) sin 2x = \(\frac{1}{2}\)
(ii) cos 3x= – \(\frac{1}{2}\)
(iii) tan \(\frac{2 x}{3}\) = √3.
Solution:
(i) Given sin 2x = \(\frac{1}{2}\)
= sin \(\frac{\pi}{6}\)
⇒ 2x = nπ + (- 1)ⁿ \(\frac{\pi}{6}\)
⇒ x = \(\frac{n \pi}{2}\) + (- 1)ⁿ \(\frac{\pi}{12}\) ∀ n ∈ I

(ii) Given cos 3x = – \(\frac{1}{2}\)
=- cos \(\frac{\pi}{3}\)
= cos (π – \(\frac{\pi}{3}\))
⇒ cos 3x = cos \(\frac{2 \pi}{3}\)
⇒ 3x = 2nπ ± \(\frac{2 \pi}{3}\)
⇒ x = \(\frac{2 n \pi}{3} \pm \frac{2 \pi}{9}\) ∀ n ∈ I

(iii) Given tan \(\frac{2 x}{3}\) = √3
⇒ \(\tan \frac{2 x}{3}=\tan \frac{\pi}{3}\)
⇒ \(\frac{2 x}{3}\) = nπ + \(\frac{\pi}{3}\) ∀ n ∈ I
⇒ x = \(\frac{3 n \pi}{2}+\frac{\pi}{2}\) ; ∀ n ∈ I

Question 7.
(i) sec 3x = – √2
(ii) cot 4x = – 1
(iii) cosec 3x = – \(\frac{2}{\sqrt{3}}\)
Solution:
(i) Given sec 3x = – √2
⇒ cos 3x = \(-\frac{1}{\sqrt{2}}=-\cos \frac{\pi}{4}\)
= cos (π – \(\frac{\pi}{4}\))
⇒ cos 3x = cos \(\frac{3 \pi}{4}\)
⇒ 3x = 2nπ ± \(\frac{3 \pi}{4}\) ∀ n ∈ I
⇒ x = \(\frac{2 n \pi}{3} \pm \frac{\pi}{4}\) ; n ∈ I

(ii) Given cot 4x = – 1
⇒ tan 4x = – 1
= – tan \(\frac{\pi}{4}\)
= tan (π – \(\frac{\pi}{4}\))
⇒ tan 4x = tan \(\frac{3 \pi}{4}\)
⇒ 4x = nπ + \(\frac{3 \pi}{4}\)
⇒ x = \(\frac{n \pi}{4}+\frac{3 \pi}{16}\) ∀ n ∈ I

(iii) cosec 3x = – \(\frac{2}{\sqrt{3}}\)
⇒ sin 3x = – \(\frac{\sqrt{3}}{2}\)
= – sin \(\frac{\pi}{3}\)
= sin (π + \(\frac{\pi}{3}\))
⇒ sin 3x = sin \(\frac{4 \pi}{3}\)
⇒ 3x = nπ + (- 1)ⁿ \(\frac{4 \pi}{3}\) ∀ n ∈ I
⇒ x = \(\frac{n \pi}{3}+(-1)^n \frac{4 \pi}{9}\) ∀ n ∈ I

Question 8.
(i) cos 3x = cos x
(ii) sin 9x sin x.
Solution:
(i) Given, cos 3x = cos x
⇒3x = 2nπ ± x ∀ n ∈ I
⇒ 3x = 2nπ + x ∀ n ∈ I
or 3x = 2nπ – x ∀ n ∈ I
⇒ x = nπ or x = \(\frac{n \pi}{2}\) ∀ n ∈ I
Also, we note that, the solutions
X = nπ ∀ n ∈ I
i.e. x = π, 2π, 3π are included in the solutions
∴ x = \(\frac{n \pi}{2}\) when n = 2, 4, 6, ……………….
Thus, the required solutions are x = \(\frac{n \pi}{2}\) ∀ n ∈ I

(ii) Given, sin 9x = sin x
⇒ 9x = nπ + (- 1)ⁿ x ∀ n ∈ I

Case – I.
When n be even integer
i.e. n = 2m
9x = 2mπ + x
⇒ x = \(\frac{m \pi}{4}\) ∀ n ∈ I

Case – II :
When n be on odd integer
i.e. n = 2m + 1
9x = (2m + 1) π – x
⇒ x = (2m + 1) \(\frac{\pi}{10}\) ∀ n ∈ I
Hence, x = \(\frac{m \pi}{4}\), (2m + 1) \(\frac{\pi}{10}\) ∀ n ∈ I

Aliter :
Given sin 9x – sin x = 0
⇒ 2 cos \(\left(\frac{9 x+x}{2}\right)\) sin \(\left(\frac{9 x-x}{2}\right)\) = 0
⇒ 2 cos 5x sin 4x = 0
either cos 5x = 0 or sin 4x = 0
⇒ 5x = (2n + 1) \(\frac{\pi}{2}\) or 4x = nπ
⇒ x = (2n + 1) \(\frac{\pi}{10}\)
or x = \(\frac{n \pi}{4}\) ∀ n ∈ I
Hence the required solutions are (2n + 1)\(\frac{\pi}{10}\) and nπ, where n ∈ I

Question 9.
(i) tan 3x = cot x
(ii) tan x + cot 3x = 0
Solution:
(i) Given tan 3x = cot x
= tan (\(\frac{\pi}{2}\) – x)
⇒ 3x = nπ + \(\frac{\pi}{2}\) – x
⇒ 4x = nπ + \(\frac{\pi}{2}\)
⇒ x = \(\frac{n \pi}{4}+\frac{\pi}{8}\), where n ∈ I
which ¡s the required general soln.

(ii) Given tan x + cot 3x = 0
⇒ tan x = – cot 3x
= tan (\(\frac{\pi}{2}\) + 3x)
⇒ x = nπ + \(\frac{\pi}{2}\) + 3x
[∵ tan θ = tan α
⇒ θ = nπ + α]
⇒ – 2x = nπ + \(\frac{\pi}{2}\)
⇒ x = \(-\frac{n \pi}{2}-\frac{\pi}{4}\)
⇒ x = \(\frac{m \pi}{2}-\frac{\pi}{4}\) ; where m ∈ I

Question 10.
(i) 2 sin x cos x = sin x
(ii) tan² x – √3 tan x = 0.
Solution:
(i) Given 2 sin x cos x = sin x
⇒ sin x (2 cos x – 1) = 0
either sin x = 0 or 2 cos x – 1 = 0
⇒ x = nπ or x = \(\frac{1}{2}\), n ∈ I
⇒ x = nπ or cos x = cos \(\frac{\pi}{3}\), n ∈ I
⇒ x = nπ or x = 2nπ ± \(\frac{\pi}{3}\) ; n ∈ I
Hence, the required solutions are nπ, 2nπ ± \(\frac{\pi}{3}\); n ∈ I

(ii) tan² x – √3 tan x = 0
⇒ tan x (tan x – √3) = 0
either tan x = 0 or tan x = √3
x = nπ or tan x = tan \(\frac{\pi}{3}\) ; n ∈ I
x = nπ or x = nπ + \(\frac{\pi}{3}\) ∀ n ∈ I
Hence the required solutions are nit,Hence the required solutions are nπ, nπ + \(\frac{\pi}{3}\); where n ∈ I

Question 11.
(i) sin x = tan x
(ii) sec² x = 1 + tan x.
Solution:
(i) Given sin x = tan x
⇒ sin x = \(\frac{\sin x}{\cos x}\)
⇒ sin x cos x – sin x = 0 [∵ cos x ≠ 0]
⇒ sin x (cos x – 1) = 0
⇒ sin x = 0 or cos x = 1 = cos 0
⇒ x = – nπ or x = 2nπ ± 0 ∀ n ∈ I
⇒ x = nπ or x = 2nπ ∀ n ∈ I
We note that the solutions x = 2nπ, where n ∈ I
i.e. x = 0, 2,π 4π, 6π are included in x = mπ
For n = 0, 2, 4, ……………….
Hence the required solution be nπ where n ∈ I

(ii) Given sec² x = 1 + tan x
⇒ 1 + tan² x = 1 + tan x
⇒ tan x (tan x – 1) = 0
either tan x = 0 or tan x = 1
⇒ x = nπ or tan x = tan \(\frac{\pi}{4}\)
⇒ x = nπ or x = nπ + \(\frac{\pi}{4}\) ∀ n ∈ I
Thus nπ, nπ + \(\frac{\pi}{4}\) ; n ∈ I are the required solutions of given equation.

Question 12.
(i) cosec² x = 2 cot x
(ii) tan x + cot x = 2.
Solution:
(i) Given, cosec² x = 2 cot x
⇒ 1 + cot² x – 2 cot x = 0
⇒ (cot x – 1)² = 0
⇒ cot x = 1
⇒ tan x = 1
⇒ tan x = tan \(\frac{\pi}{4}\)
⇒ x = nπ + \(\frac{\pi}{4}\) where n ∈ I

(ii) Given tan x + cot x = 2
⇒ tan x + \(\frac{1}{\tan x}\) = 2
⇒ tan² x – 2 tan x + 1 = 0
[∵ tan x ≠ 0]
⇒ (tan x – 1)² = 0
⇒ tan x = 1 = tan \(\frac{\pi}{4}\)
⇒ x = nπ + \(\frac{\pi}{4}\) ; where n ∈ I

Question 13.
(i) tan x + cot x = 2 sec x
(ii) cos 3x + cos x – 2 cos 2x = 0.
Solution:
(i) Given tan x + cot x = 2 sec x
⇒ \(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{2}{\cos x}\)
⇒ \(\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x}=\frac{2}{\cos x}\)
⇒ \(\frac{1}{\sin x \cos x}=\frac{2}{\cos x}\)
⇒ 2 sin x = 1
[∵ cos x ≠ 0]
⇒ sin x = \(\frac{1}{2}\)
= sin \(\frac{\pi}{6}\)
⇒ x = nπ + (- 1)ⁿ \(\frac{\pi}{6}\) ∀ n ∈ I

(ii) Given cos 3x + cos x – 2 cos 2x = 0
⇒ 2 cos \(\left(\frac{3 x+x}{2}\right)\) cos \(\left(\frac{3 x-x}{2}\right)\) – 2 cos 2x = 0
[∵ cos C + cos D = 2 cos \(\left(\frac{\mathrm{C}+\mathrm{D}}{2}\right)\) cos \(\left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)]
⇒ 2cos 2x cos x – 2 cos 2x = 0
⇒ 2 cos 2x (cos x – 1) = 0
either cos 2x = 0 or cos x – 1 = 0
i.e. 2x = (2n +1) \(\frac{\pi}{2}\) or cos x = 1 = cos 0
⇒ x = (2n + 1) \(\frac{\pi}{4}\) or x = 2nπ ; where n ∈ I

Question 14.
(i) sin 2x + sin 4x + sin 6x = 0
(ii) sin x – sin 2x + sin 3x = 0
Solution:
(i) Given, sin 2x + sin 4x + sin 6x = 0
⇒ (sin 6x + sin 2x) + sin 4x = 0
⇒ 2 sin \(\left(\frac{6 x+2 x}{2}\right)\) cos \(\left(\frac{6 x-2 x}{2}\right)\) + sin 4x = 0
⇒ 2 sin 4x cos 2x + sin 4x = 0
⇒ sin 4x (2 cos 2x + 1) = 0
either sin 4x = 0
or cos 2x = – \(\frac{1}{2}\) = – cos \(\frac{\pi}{3}\)
⇒ 4x = nπ or cos 2x = + cos (π – \(\frac{\pi}{3}\))
⇒ 4x = nπ or 2x = 2nπ ± \(\frac{2 \pi}{3}\) ∀ n ∈ I
⇒ x = \(\frac{n \pi}{4}\) or x = nπ ± \(\frac{\pi}{3}\) ; n ∈ I
Thus the required solutions are \(\frac{n \pi}{4}\) or nπ ± \(\frac{\pi}{3}\), n ∈ I

(ii) Given sin x – sin 2x + sin 3x = 0
⇒ (sin 3x + sin x) – sin 2x = 0
⇒ 2 sin \(\left(\frac{3 x+x}{2}\right)\) cos \(\left(\frac{3 x-x}{2}\right)\) – sin 2x = 0
⇒ 2 sin 2x cos x – sin 2x = 0
⇒ sin 2x (2 cos x – 1) = 0
either sin 2x = 0
or cos x = \(\frac{1}{2}\)
= cos \(\frac{\pi}{3}\)
⇒ 2x = nπ or x = 2nπ ± \(\frac{\pi}{3}\) ; n ∈ I
Thus, required solutions are \(\frac{n \pi}{2}\) or 2nπ ± \(\frac{\pi}{3}\) ; n ∈ I.

Question 15.
(i) 7 cos² x + 3 sin² x = 4
(ii) Zsin2x=3cosx.
Solution:
(i) Given eqn. be,
7 cos² x + 3 sin² x = 4
⇒ 7 cos² x + 3 (1 – cos² x) = 4
⇒ 4 cos² x = 1
⇒ cos² x = \(\frac{1}{4}=\left(\frac{1}{2}\right)^2\)
= cos² \(\frac{\pi}{3}\)
⇒ x = nπ ± \(\frac{\pi}{3}\), where n ∈ I
[∵ cos² x = cos² α
⇒ x = nπ ± α where n ∈ I]

(ii) Given, 2 sin² x = 3 cos x
⇒ 2 (1 cos² x) = 3 cos x
⇒ 2 cos² x + 3 cos x – 2 = 0
⇒ cos x = \(\frac{-3 \pm \sqrt{9+16}}{4}\)
= \(\frac{-3 \pm 5}{4}\)
= \(\frac{1}{2}\), – 2
Since cos x = – 2 is not possible
[∵ – 1 ≤ cos x ≤ 1]
Thus, cos x = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
⇒ x = 2nπ ± \(\frac{\pi}{3}\) ; n ∈ I

Question 16.
(i) 4 cos² x – 4 sin x – 1 = 0
(ii) 2 sin² x + √3 cos x + 1 = 0
Solution:
(i) Given 4 cos² x – 4 sin x – 1 = 0
⇒ 4(1 – sin² x) – 4 sin x – 1 = 0
⇒ 4 sin² x + 4 sin x – 3 = 0
⇒ sin x = \(\frac{-4 \pm \sqrt{16+48}}{8}\)
= \(\frac{-4 \pm 8}{8}\)
= \(\frac{1}{2},-\frac{3}{2}\)
But sin x = – \(\frac{3}{2}\) is not possible
[∵ |sin x| ≤ 1]
Thus sin x = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
⇒ x = nπ + (- 1)ⁿ \(\frac{\pi}{6}\) ; where n ∈ I

(ii) Given, 2 sin² x + √3 cos x + 1 = 0.
⇒ 2 (1 – cos² x) + cos x + 1 = 0
⇒ 2 cos² x – √3 cos x – 3 = 0
⇒ cos x = \(\frac{\sqrt{3} \pm \sqrt{3+24}}{4}\)
= \(\frac{\sqrt{3} \pm 3 \sqrt{3}}{4}\)
cos x = \(\frac{4 \sqrt{3}}{4}, \frac{-2 \sqrt{3}}{4}\)
cos x = √3, \(\frac{-\sqrt{3}}{2}\)
But cos x = √3 is not possible.
[∵ |cos x| ≤ 1]
Thus, cos x = \(\frac{-\sqrt{3}}{2}\)
= – cos \(\frac{\pi}{6}\)
= cos (π – \(\frac{\pi}{6}\))
⇒ cos x = cos \(\frac{5 \pi}{6}\)
⇒ x = 2nπ ± \(\frac{5 \pi}{6}\) ; n ∈ I

Question 17.
(i) cos x + sin x = 1
(ii) cos x – sin x+ 1 = 0
(iii) √3 cos x – sin x = 1
Solution:
(i) Given cos x + sin x = 1 ………………(1)
which is of the form a cos x + b sin x = C
Dividing throughout eqn. (1) by \(\sqrt{a^2+b^2}\)
i.e. \(\sqrt{1^2+1^2}\) i.e. √2 ; we have

(ii) Given cos x – sin x + 1 = 0
⇒ sin x – cos x = 1
⇒ cos x – sin x = – 1 …………………..(1)
which is of the form a cos x + b sin x = c
So dividing throughout eqn. (1) by
\(\sqrt{1^2+(-1)^2}\) i.e. by √2 ; we have

(iii) Given eqn. be,
√3 cos x – sin x = 1
⇒ \(\frac{\sqrt{3}}{2} \cos x-\frac{1}{2} \sin x=\frac{1}{2}\)
⇒ cos x cos \(\frac{\pi}{6}\) – sin x sin \(\frac{\pi}{6}\) = \(\frac{1}{2}\)
⇒ cos (x + \(\frac{\pi}{6}\)) = cos \(\frac{\pi}{3}\)
⇒ x + \(\frac{\pi}{6}\) = 2nπ ± \(\frac{\pi}{3}\)
[∵ cos x = cos α
⇒ x = 2nπ ± α where n ∈ I]
⇒ x = 2nπ ± \(\frac{\pi}{3}-\frac{\pi}{6}\)
⇒ x = 2nπ + \(\frac{\pi}{6}\), 2nπ – \(\frac{\pi}{2}\), where n ∈ I

Question 18.
(i) √2 sec x + tan x = 1
(ii) cosec x = cot x + √3
Solution:
(i) Given √2 sec x + tan x = 1
⇒ \(\frac{\sqrt{2}}{\cos x}+\frac{\sin x}{\cos x}\) = 1
⇒ √2 + sin x = cos x ; cos x ≠ 0
⇒ cos x – sin x = √2, cos x ≠ 0
which is of the form a cos x + b sin x = c
So dividing throughout eqn. (1) by \(\sqrt{1^2+(-1)^2}\) i.e. by √2 ; we have
\(\frac{1}{\sqrt{2}}\) cos x – \(\frac{1}{\sqrt{2}}\) sin x = 1
⇒ cos \(\frac{\pi}{4}\) cos x – sin \(\frac{\pi}{4}\) sin x = 1
⇒ cos (x + \(\frac{\pi}{4}\)) = cos 0
⇒ x + \(\frac{\pi}{4}\) = 2nπ ∀ n ∈ I
⇒ x = 2nπ – \(\frac{\pi}{4}\) ∀ n ∈ I

(ii) cosec x = cot x + √3
⇒ \(\frac{1}{\sin x}=\frac{\cos x}{\sin x}\) + √3
⇒ 1 = cos x + √3 sin x, sin x ≠ 0 ……………….(1)
which n of the form a cos x + b sin x = c
So dividing throughout eqn. (1) by \(\sqrt{1^2+(\sqrt{3})^2}\) i.e. 2 ; we have
i.e. \(\frac{1}{2}\) cos x + \(\frac{\sqrt{3}}{2}\) sin x = \(\frac{1}{2}\)
⇒ cos \(\frac{\pi}{3}\) cos x + sin \(\frac{\pi}{3}\) sin x = \(\frac{1}{2}\)
⇒ cos (x – \(\frac{\pi}{3}\)) = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
⇒ x – \(\frac{\pi}{3}\) = 2nπ ± \(\frac{\pi}{3}\) ; n ∈ I
⇒ x = 2nπ ± \(\frac{\pi}{3}\) + \(\frac{\pi}{3}\) ∀ n ∈ I
i.e. x = 2nπ + \(\frac{\pi}{3}\), 2nπ where n ∈ I
But sin (2nπ) = 0 so we reject these values.
Hence the required solutions are
x = 2nπ + \(\frac{2 \pi}{3}\) ∀ n ∈ I