Question

If $\sqrt{3}\cot x-\text{cosec}x=1$, then $x$ can be
(where $n\in Z$)

• A. $2n\pi +\frac{\pi }{6}$
• B. $2n\pi -\frac{\pi }{2}$
• C. $2n\pi +\frac{\pi }{2}$
• D. $2n\pi -\frac{\pi }{6}$

Answer 1

Answer: (A), (B)

The correct options are
A $2n\pi +\frac{\pi }{6}$
B $2n\pi -\frac{\pi }{2}$

$\sqrt{3}\cot x-\text{cosec}x=1$
$\Rightarrow \sqrt{3}\frac{\cos x}{\sin x}-\frac{1}{\sin x}=1$
$\Rightarrow \sqrt{3}\cos x-1=\sin x$
$\Rightarrow \sqrt{3}\cos x-\sin x=1...\left(1\right)$
Dividing both sides of $\left(1\right)$ by $\sqrt{(\sqrt{3}{)}^2+(-1{)}^2}$ i.e., by $2$, we get
$\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x=\frac{1}{2}$
$\Rightarrow \cos x\cos \frac{\pi }{6}-\sin x\sin \frac{\pi }{6}=\frac{1}{2}$
$\Rightarrow \cos (x+\frac{\pi }{6})=\cos \frac{\pi }{3}$
$\Rightarrow (x+\frac{\pi }{6})=2n\pi \pm \frac{\pi }{3}$
$x=2n\pi +\frac{\pi }{6}\text{or}2n\pi -\frac{\pi }{2},n\in Z.$