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

11th Grade > Mathematics

TRIGONOMETRIC EQUATIONS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

Find the general solution of $c o s^{2} x - 1 = 0$

$x = n π$
$x = 2 n π + \frac{π}{2}$
$x = n π + \frac{π}{2}$
$x = 2 n π$

Discuss Question

Answer: Option A. ->

x = n π

:
A

$c o s^{2} x - 1 = 0 1 - c o s^{2} x = 0 s i n^{2} x = 0 s i n x = 0 x = n π$

Question 2.

What is the principal solution of 2 sin x = 1?

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

Discuss Question

Answer: Option A. ->

x = \frac{π}{6}

:
A

2 sin x =1
$s i n x = \frac{1}{2} x = \frac{π}{6}$

Question 3.

Find the principal solution of $t a n^{2} x + (\sqrt{3} - 1) t a n x - \sqrt{3} = 0$

$x = \frac{π}{4}$
$x = \frac{π}{3}$
$x = - \frac{π}{4}$
$x = - \frac{π}{3}$

Discuss Question

Answer: Option A. ->

x = \frac{π}{4}

:
A and D

$t a n^{2} x + (\sqrt{3} - 1) t a n x - \sqrt{3} = 0 t a n^{2} x - t a n x + \sqrt{3} t a n x - \sqrt{3} = 0 t a n x (t a n x - 1) + \sqrt{3} (t a n x - 1) = 0 (t a n x + \sqrt{3}) (t a n x - 1) = 0 t a n x = - \sqrt{3} o r t a n x = 1 x = - \frac{π}{3} o r x = \frac{π}{4}$

Question 4.

Find the principal solutions of $2 s i n^{2} x - s i n x - 1 = 0$

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

Discuss Question

Answer: Option B. ->

x = \frac{π}{2}

:
B and C

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

Question 5.

Find the principal solution of $\frac{1 + c o s x}{c o s x} = 2$

$x = 0$
$x = \frac{π}{2}$
$x = π$
none of these

Discuss Question

Answer: Option A. ->

x = 0

:
A

$\frac{1 + c o s x}{c o s x} = 2 1 + c o s x = 2 c o s x c o s x = 1 x = 0$

Question 6.

Find the principal solution of $\sqrt{\frac{1 + 2 s i n x}{2}} = 1$

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

Discuss Question

Answer: Option A. ->

x = \frac{π}{6}

:
A

$\sqrt{\frac{1 + 2 s i n x}{2}} = 1$
Squaring, we get
$\frac{1 + 2 s i n x}{2} = 1 1 + 2 s i n x = 2 2 s i n x = 1 s i n x = \frac{1}{2} x = \frac{π}{6}$

Question 7.

Find the general solutions of cos 4x = cos 2x

$x = n π$
$x = n π \pm \frac{π}{6}$
$x = n π \pm \frac{π}{3}$
$x = n π \pm \frac{π}{4}$

Discuss Question

Answer: Option A. ->

x = n π

:
A and C

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

Question 8.

Find the principal solutions of cos 3x - cos 2x + cos x = 0

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

Discuss Question

Answer: Option A. ->

x = \frac{π}{4}

:
A, B, and D

$c o s 3 x - c o s 2 x + c o s x = 0 4 c o s^{3} x - 3 c o s x - 2 c o s^{2} x + 1 + c o s x = 0 4 c o s^{3} x - 2 c o s^{2} x - 2 c o s x + 1 = 0 2 c o s^{2} x (2 c o s x - 1) - 1 (2 c o s x - 1) = 0 (2 c o s^{2} x - 1) (2 c o s x - 1) = 0 c o s^{2} x = \frac{1}{2} o r c o s x = \frac{1}{2} c o s x = \frac{1}{2} \Rightarrow x = \frac{π}{3} c o s^{2} x = \frac{1}{2} \Rightarrow c o s x = \pm \frac{1}{\sqrt{2}} \Rightarrow x = \frac{π}{4} o r x = \frac{3 π}{4}$