Trigonometric Equations(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

TRIGONOMETRIC EQUATIONS MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21.

What is the principal solution of $4 c o s^{2} x - 8 c o s x + 3 = 0 ?$

$x = \frac{π}{6}$
$x = \frac{π}{4}$
$x = \frac{π}{3}$
$x = \frac{π}{2}$

Discuss Question

Answer: Option C. ->

x = \frac{π}{3}

:
C

$4 c o s^{2} x - 8 c o s x + 3 = 0 4 c o s^{2} x - 8 c o s x + 4 = 1 (2 c o s x - 2)^{2} = 1 2 c o s x - 2 = \pm 1 2 c o s x = 3 - n o t p o s s i b l e 2 c o s x = 1 c o s x = \frac{1}{2} x = \frac{π}{3}$

Question 22.

What are the principal solutions of the equation
$2 s i n^{2} x + (2 - \sqrt{3}) s i n x - \sqrt{3} = 0 ?$

$x = \frac{π}{3}$
$x = - \frac{π}{3}$
$x = \frac{π}{2}$
$x = - \frac{π}{2}$

Discuss Question

Answer: Option A. ->

x = \frac{π}{3}

:
A and D

$2 s i n^{2} x + (2 - \sqrt{3}) s i n x - \sqrt{3} = 0 2 s i n^{2} x + 2 s i n x - \sqrt{3} s i n x - \sqrt{3} = 0 2 s i n x (s i n x + 1) - \sqrt{3} (s i n x + 1) = 0 (2 s i n x - \sqrt{3}) (s i n x + 1) = 0 s i n x = \frac{\sqrt{3}}{2} o r s i n x = - 1 x = \frac{π}{3} o r x = - \frac{π}{2}$

Question 23.

Find the principal solution of $3 c o s^{2} θ - c o s 2 θ = 1$

$θ = 0$
$θ = - \frac{π}{2}$
$θ = \frac{π}{2}$
$θ = π$

Discuss Question

Answer: Option C. ->

θ = \frac{π}{2}

:
C

$3 c o s^{2} θ - c o s 2 θ = 1 3 c o s^{2} θ - (2 c o s^{2} θ - 1) = 1 c o s^{2} θ = 0 c o s θ = 0 θ = \frac{π}{2}$

Question 24.

What is the principal solution of $s i n x + \sqrt{s i n x} = 0 ?$

$x = - \frac{π}{2}$
$x = 0$
$x = \frac{π}{2}$
$x = π$

Discuss Question

Answer: Option B. ->

x = 0

:
B

$s i n x + \sqrt{s i n x} = 1 L e t \sqrt{s i n x} = t ∴ t^{2} + t = 0 t^{2} = - t t = 0 o r t = - 1 \sqrt{s i n x} = 0 o r \sqrt{s i n x} = - 1 \sqrt{s i n x} = - 1 - n o t p o s s i b l e s i n x = 0 x = 0$

Question 25.

Find the principal solution of cos 2x= sin x

$x = \frac{π}{6}$
$x = \frac{π}{2}$
$x = - \frac{π}{6}$
$x = - \frac{π}{2}$

Discuss Question

Answer: Option A. ->

x = \frac{π}{6}

:
A and D

$c o s 2 x = s i n x 1 - 2 s i n^{2} x = s i n x 2 s i n^{2} x + s i n x - 1 = 0 L e t s i n x = t 2 t^{2} + t - 1 = 0 t = \frac{- 1 \pm \sqrt{1 - 4 \times 2 \times - 1}}{2 \times 2} t = \frac{- 1 \pm 3}{4} s i n x = \frac{1}{2} o r s i n x = - 1 x = \frac{π}{6} o r x = - \frac{π}{2}$

Question 26.

The general solution of $t a n x = t a n α, α ϵ (- \frac{π}{2}, \frac{π}{2})$ is

$x = n π + (- 1)^{n} α, n ϵ Z$
$x = n π + α, n ϵ Z$
$x = 2 n π \pm α, n ϵ Z$
None of these

Discuss Question

Answer: Option B. ->

x = n π + α, n ϵ Z

:
B

The general solution of $t a n x = t a n α, α ϵ (- \frac{π}{2}, \frac{π}{2})$ is
$x = n π + α, n ϵ Z$

Question 27.

Find the principal solution of sin x + sin 3x +sin 5x = 0

$x = 0$
$x = \frac{π}{6}$
$x = \frac{π}{3}$
$x = \frac{π}{2}$

Discuss Question

Answer: Option A. ->

x = 0

:
A and C

$s i n x + s i n 3 x + s i n 5 x = 0 (s i n x + s i n 5 x) + s i n 3 x = 0 2 s i n 3 x c o s 2 x + s i n 3 x = 0 s i n 3 x (2 c o s 2 x + 1) = 0 s i n 3 x = 0 o r c o s 2 x = - \frac{1}{2} 3 x = 0 o r 2 x = \frac{2 π}{3} x = 0 o r x = \frac{π}{3}$

Question 28.

Find the general solutions of $2 c o s^{2} x + \sqrt{3} c o s x = 0$

$x = n π$
$x = 2 n π \pm \frac{π}{2}$
$x = 2 n π \pm \frac{5 π}{6}$
$x = 2 n π \pm \frac{π}{6}$

Discuss Question

Answer: Option B. ->

x = 2 n π \pm \frac{π}{2}

:
B and C

$2 c o s^{2} x + \sqrt{3} c o s x = 0 c o s x (2 c o s x + \sqrt{3}) = 0 c o s x = 0 o r c o s x = - \frac{\sqrt{3}}{2} x = 2 n π \pm \frac{π}{2} o r x = 2 n π \pm \frac{5 π}{6}$

Question 29.

Find the general solution of $s i n x + \sqrt{2} = - s i n x$

$x = n π + (- 1)^{n + 1} \frac{π}{4}$
$x = n π + (- 1)^{n} \frac{π}{4}$
$x = n π + (- 1)^{n} \frac{π}{3}$
$x = n π + (- 1)^{n} \frac{3 π}{4}$

Discuss Question

Answer: Option A. ->

x = n π + (- 1)^{n + 1} \frac{π}{4}

:
A

$s i n x + \sqrt{2} = - s i n x 2 s i n x = - \sqrt{2} s i n x = - \frac{1}{\sqrt{2}} x = n π + (- 1)^{n} (- \frac{π}{4}) x = n π + (- 1)^{n + 1} \frac{π}{4}$