# CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(c)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 4 Trigonometric Functions Ex 4(c) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Exercise 4(c)

Question 1.

Fill in the blanks choosing the correct answer from the brackets.

(i) The number of solutions of  2 sin θ – 1 = 0 is__________. (one, two, infinite)
Solution:
Infinite

(ii) If cos α = cos β, then α + β = ____________. (0, π, 2π)
Solution:

(iii) The number of solution(s) of 2 sin θ + 1 = 0 is__________. (zero, two, infinite)
Solution:
Zero

(iv) If tan θ = tan α and 90° < α < 180°, then θ can be in ____________quadrant. (1st, 3rd, 4th)
Solution:
4th

(v) If tan x. tan 2x. tan 7x = tan x + tan 2x + tan 7x, then x = _____________. ($$\frac{\pi}{4}, \frac{\pi}{5}, \frac{\pi}{10}$$)
Solution:
$$\frac{\pi}{10}$$

(vi) For_____________value of θ, sin θ + cos θ = √2. ($$\frac{\pi}{4}, \frac{\pi}{2}, \frac{\pi}{3}$$)
Solution:
$$\frac{\pi}{4}$$

(vii) The number of values of x for which cos2 x = 1 and x2 ≤ 4 is______________. (1, 2, 3)
Solution:
1

(viii) In the 1st quadrant the solution of tan2 θ = 3 is_____________. ($$\frac{\pi}{2}, \frac{\pi}{3}, \frac{\pi}{4}$$)
Solution:
$$\frac{\pi}{3}$$

(ix) The least positive value of θ for which 1 + tan θ = 0 and √2 cos θ + 1 = 0 is___________. ($$\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}$$)
Solution:
$$\frac{3 \pi}{4}$$

(x) the least positive value of x for which tan 3x = tan x is______________. ($$\frac{\pi}{2}, \frac{\pi}{3}, \pi$$)
Solution:
$$\frac{\pi}{2}$$

Question 2.
Find the principal solution of the following equations:
(i) sin θ = sin 2θ
Solution:
sin θ = sin 2θ
or, 2θ = nπ + (-1)n θ
or, 2π – (-1)n θ = nπ
or, θ = $$\frac{n \pi}{2-(-1)^n}$$
when n = 0, θ = 0
when n = 1, θ = $$\frac{\pi}{3}$$
when n = 2, θ = 2π
when n = 3, θ = π
when n = 4, θ = 4π
when n = 5, θ = $$\frac{5 \pi}{3}$$
∴ The principal solution are 0, $$\frac{\pi}{3}$$, π, $$\frac{5 \pi}{3}$$

(ii) √3 sin θ – cos θ = 2
Solution:
√3 sin θ – cos θ = 2
or, $$\frac{\sqrt{3}}{2}$$ sin θ – 1/2 cos θ = 1

which is the only principal solution.

(iii) cos2 θ + sin θ + 1 = 0
Solution:
cos2 θ + sin θ + 1 = 0
or, 1 – sin2 θ + sin θ + 1 = 0
or,  sin2 θ – sin θ + 2 = 0
or, sin2 θ – 2 sin θ + sin θ – 2 = 0
or, sinθ(sinθ – 2) + (sinθ – 2) = 0
or, (sinθ – 2) (sinθ + 1) = 0
∴ sinθ = 2, sinθ = – 1
= sin $$\left(-\frac{3 \pi}{2}\right)$$ or, θ = – $$\frac{\pi}{2}$$, $$\frac{3 \pi}{2}$$
∴ The principal solution is $$\frac{3 \pi}{2}$$.

(iv) sin 4x + sin 2x = 0
Solution:

(v) sin x + cos x = $$\frac{1}{\sqrt{2}}$$
Solution:

Question 3.
Find the general solution of the following equations:
(i) cos 2x = θ
Solution:
cos 2x = θ
or, 2x = (2n + 1)$$\frac{\pi}{2}$$
or, x = (2n + 1)$$\frac{\pi}{4}$$, n∈Z

(ii) sin(x° + 40°) = $$\frac{1}{\sqrt{2}}$$
Solution:

(iii) sin 5θ = sin 3θ
Solution:
sin 5θ = sin 3θ
or, 5θ = nπ + (-1)n
or, 5θ – (-1)n 3θ = nπ
or, θ[5 – (-1)n3] = nπ
or, θ = $$\frac{n \pi}{5-(-1)^n 3}$$

(iv) tan ax = cot bx
Solution:

(v) tan2 3θ = 3
Solution:

Question 4.
Solve the following:
(Hints : cos x ≠ 0 and sin2 x- sin x + 1/2 = 0)
(i) tan2 x + sec2 x = 3
Solution:
tan2 x + sec2 x = 3
or, tan2 x + 1 + tan2 x = 3
or, 2tan2 x = 2
or, tan2 x = 1
or, tan x = ± 1 = tan $$\left(\pm \frac{\pi}{4}\right)$$
∴ x = nπ ± $$\frac{\pi}{4}$$

(ii) 4 sin2 x + 6 cos2 x = 5
Solution:
4 sin2 x + 6 cos2 x = 5
or, 4 sin2 x + 6(1 – sin2 x) = 5
or, 4 sin2 x + 6 – 6 sin2 x = 5
or, 6- 2 sin2 x = 5
or, 2 sin2 x = 1
or, sin2 x = 1/2
or, sin x = ± $$\frac{1}{\sqrt{2}}$$ = sin $$\left(\pm \frac{\pi}{4}\right)$$
or, x = nπ + (-1)n $$\left(\pm \frac{\pi}{4}\right)$$

(iii) 3 sin x + 4 cos x = 5
Solution:

(iv) 3 tan x + cot x = 5 cosec x
Solution:

(v) cos x + √3 sin x = √2
Solution:

(vi) sin 3x – 2 cos2 x = 0
Solution :
sin 2x – 2 cos2 x = 0
or, 2 sin x cos x – 2 cos2 x = 0
or, 2 cos x(sin x – cos x) = 0
∴ cos x = 0, sin x = cos x
∴ x = (2n + 1)$$\frac{\pi}{2}$$, tan x = 1 = tan $$\frac{\pi}{4}$$
or, x = nπ + $$\frac{\pi}{4}$$

(vii) sec θ + tan θ = √3
Solution:

(viii) cos 2θ – cos θ = sin θ – sin 20
Solution:

(ix) sin θ + sin 2θ + sin 3θ + sin 4θ = 0
Solution:

(x) cos 2x° + cos x° – 2 = 0
Solution:
cos 2x° + cos x° – 2 = 0
or, 2 cos2 x° – 1 + cos x° – 2 = 0
or, 2 cos2 x° + cos x° – 3 = 0
or 2 cos2 + 3cos x° – 2cos x°- 3 = 0
or, cos x°(2 cos x° + 3) – 1(2 cos x° + 3) = 0
or, (2 cos x° + 3)(cos x° – 1) = 0
∴ cos x° = 1 = cos 0°
∴ x° = 2nπ ± 0 = 2nπ
or, $$\frac{\pi x}{180}$$ = 2nπ
or, x = 360 n
Again 2 cos x° + 3 = 0
⇒ cos x° = – 3/2 which has no solution.
Hence x = 360 n.

(xi) tan θ + tan 2θ = tan 3θ
Solution:

(xii) tan θ + tan ($$\theta+\frac{\pi}{3}$$) + tan ($$\theta+\frac{2\pi}{3}$$) = 3
Solution:

(xiii) cot2 θ – tan2 θ = 4 cot 2θ
Solution:

(xiv) cos 2θ = $$(\sqrt{2}+1)\left(\cos \theta-\frac{1}{\sqrt{2}}\right)$$
Solution:

(xv) sec θ – 1 = $$(\sqrt{2}-1)$$ tan θ
Solution:

⇒ θ = 2nπ + $$\frac{\pi}{4}$$

(xvi) 3cot2 θ – 2 sin θ = 0
Solution:

(xvii) 4 cos x. cos 2x . cos 3x = 1
Solution:
4 cos x cos 2x cos 3x = 1
⇒ 2 cos x cos 2x. 2 cos 3x = 1
⇒ (cos 3x + cos x) 2 cos 3x = 1
⇒ 2 cos2 3x + 2 cos 3x cos x = 1
⇒ 2 cos2 3x – 1 + cos 4x + cos 2x = 0
⇒ cos 6x + cos 4x + cos 2x = 0
⇒ cos 6x + cos 2x + cos 4x = 0
⇒ 2 cos 4x cos 2x + cos 4x = 0
⇒ cos 4x (2 cos 2x + 1) = 0
⇒ cos 4x = 0, cos 2x = – 1/2
cos 4x = 0 ⇒ 4x = (2n + 1) $$\frac{\pi}{2}$$

(xviii) cos 3x – cos 2x = sin 3x
Solution:

⇒ 1 – 2 sin x cos x = y2
∴ Equation (1) reduces to
1 – 2(1 – y2) + y = 0
⇒ 2y2 + y – 1 = 0
⇒ (2y- 1) (y + 1) = 0

(xix) cos x + sin x = cos 2x + sin 2x
Solution:
cos x + sin x – cos 2x + sin 2x
[Refer (viii)]

(xx) tan x + tan 4x + tan 7x = tan x. tan 4x. tan 7x
Solution:
tan x + tan 4x + tan 7x = tan x tan 4x tan 7x
or, tan x + tan 4x
= – tan 7x + tan x tan 4x tan 7x
= – tan 7x (1 – tan x tan 4x)
or, $$\frac{\tan x+\tan 4 x}{1-\tan x \tan 4 x}$$ = – tan 7x
or, tan (x + 4x) = tan (π – 7x)
or, tan 5x = tan (π – 7x)
or, 5x = nπ + π – 7x
or, 12x = π(n + 1)
or, x = $$\frac{\pi(n+1)}{12}$$, n∈Z

(xxi) 2(sec2 θ + sin2 θ) = 5
Solution:

(xxii) $$(\cos x)^{\sin ^2 x-\frac{3}{2} \sin x+\frac{1}{2}}=1$$
Solution:
$$(\cos x)^{\sin ^2 x-\frac{3}{2} \sin x+\frac{1}{2}}=0$$
As cos x ≠ 0.
we have sin2 – $$\frac{3}{2}$$ sin x + $$\frac{1}{2}$$ = 0
∴ 2 sin2 x – 3 sin x + 1 = 0
or, 2 sin2 x – 2 sin x – sin x + 1 = 0
or, (2sin x – 1)(sin x – 1) = 0
∴ sin x = $$\frac{1}{2}$$ or, sin x = 1
But as cos x ≠ 0, we have sin x ≠ 1
∴ sin x = $$\frac{1}{2}$$ = sin $$\frac{\pi}{6}$$
∴ x = nπ + (-1)n $$\frac{\pi}{6}$$, n∈Z