Question

The solution of the equation $\cos ^2 x-2 \cos x=4 \sin x+4 \sin ^2 x(0 \leq x \leq \pi)$ 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 $
$1-\tan ^2 \frac{x}{2}-4 \tan \frac{x}{2}=2+2 \tan ^2 \frac{x}{2} \Rightarrow 3 \tan ^2 \frac{x}{2}+4 \tan \frac{x}{2}+1=0$
$\left(3 \tan \frac{x}{2}+1\right)\left(\tan \frac{x}{2}+1\right)=0 \Rightarrow \tan \frac{x}{2}=\frac{-1}{3} \text { or }-1$
$\left.\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}\right] $