Question

The number of solutions of the equation|cot x| = cot x + $\frac{1}{sinx}$ ,0 < x < 2$\pi$, is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B

1

If cot x > 0

then $\frac{1}{sinx}$ = 0 (impossible) Now if cot x < 0, then

-cot x = cot x + $\frac{1}{sinx}$

$\Rightarrow \frac{2cosx+1}{sinx}$ = 0 ⇒ cos x = - $\frac{1}{2}$

$\Rightarrow$ cos x = cos( $\frac{2\pi }{3}$)

x = 2n$\pi$ ± $\frac{2\pi }{3}$; n ∈ I

and 0 ≤ x ≤ 2$\pi$ then x = $\frac{2\pi }{3}$, $\frac{4\pi }{3}$

But cotx < 0 $\therefore$ x = $\frac{2\pi }{3}$