Question

The solution of the equation ${\cos }^{2}x-2\cos x=4\sin x+4{\sin }^{2}x(0\le x\le \pi )$ is

• A. $\pi +{\tan }^{-1}\left(\frac{-1}{2}\right)$
• B. $\pi -{\cot }^{-1}\left(\frac{1}{2}\right)$
• C. $\pi -{\tan }^{-1}2$
• D. none of these

Answer 1

Answer: (A)

We have, ${\cos }^{2}x-2\cos x=4\sin x+4{\sin }^{2}x$
${\cos }^{2}x-4{\sin }^{2}x=4\sin x+2\cos x$
$(\cos x+2\sin x)(\cos x-2\sin x-2)=0$
$\tan x=\frac{-1}{2},\cos x-2\sin x=2$
$\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}-\frac{2\cdot 2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=2$
$1-{\tan }^2\frac{x}{2}-4\tan \frac{x}{2}=2+2{\tan }^2\frac{x}{2}\Rightarrow 3{\tan }^2\frac{x}{2}+4\tan \frac{x}{2}+1=0$
$\left(3\tan \frac{x}{2}+1\right)\left(\tan \frac{x}{2}+1\right)=0\Rightarrow \tan \frac{x}{2}=\frac{-1}{3}\text{ or }-1$
$\Rightarrow \frac{x}{2}={\tan }^{-1}\left(\frac{1}{3}\right)\text{ or }\frac{-\pi }{4}\Rightarrow x=-2{\tan }^{-1}\left(\frac{1}{3}\right),\frac{-\pi }{2}]$