Question

General solution of equation $1+2\mathrm{cosec}x=-\frac{{\sec }^2\frac{x}{2}}{2}$ is

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

Answer 1

Answer: (C)

$1+2\mathrm{cosec}x=\frac{-{\sec }^2\left(\frac{x}{2}\right)}{2}$
$\Rightarrow 1+\frac{2}{\sin x}=\frac{-1}{1+\cos x}$
$\Rightarrow (2+\sin x)(1+\cos x)=-\sin x$
$\Rightarrow 2+2\cos x+\sin x+\sin x\cos x=-\sin x$
$\Rightarrow 2(\sin x+\cos x)+\sin x\cos x+2=0$
Put $\sin x+\cos x=t$
$\Rightarrow 1+2\sin x\cos x=t^2$
$\therefore 2t+\frac{t^2-1}{2}+2=0\Rightarrow t^2+4t+3=0$
$\Rightarrow t=-1,-3\Rightarrow \sin x+\cos x=-1$
$\Rightarrow \cos \left(x-\frac{\pi }{4}\right)=\frac{-1}{\sqrt{2}}=\cos \frac{3\pi }{4}$
$\Rightarrow x-\frac{\pi }{4}=2n\pi \pm \frac{3\pi }{4}$
$\Rightarrow x=2n\pi +\pi ,2n\pi -\frac{\pi }{2}$
$\Rightarrow x=2n\pi +\pi$ at which $\text{cosec}x$ is not defined
$\therefore x=2n\pi -\frac{\pi }{2}$.