Question

${\cos }^{2}x-2\cos x=4\sin x-\sin 2x$ if

• A. $\tan \left(\frac{x}{2}\right)=-\frac{1}{2}$
• B. $\tan x=-\frac{1}{2}$
• C. $\tan \left(\frac{x}{2}\right)=2+\sqrt{5}$
• D. $\tan \left(\frac{x}{2}\right)=2-\sqrt{5}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\tan x=-\frac{1}{2}$
C $\tan \left(\frac{x}{2}\right)=2+\sqrt{5}$
D $\tan \left(\frac{x}{2}\right)=2-\sqrt{5}$

$\cos \left(x\right)\left[\cos \right(x)-2]=2\sin \left(x\right)[2-\cos (x\left)\right]$
Or
$\cos \left(x\right)\left[\cos \right(x)-2]+2\sin \left(x\right)\left[\cos \right(x)-2]=0$
Or
$\left(\cos \right(x)-2)\left(\cos \right(x)+2\sin (x\left)\right)=0$
Now
$\cos \left(x\right)=2$ is not possible.
Hence
$\cos \left(x\right)=-2\sin \left(x\right)$
Or
$\tan \left(x\right)=\frac{-1}{2}$.
Or
$\frac{2\tan \left(\frac{x}{2}\right)}{1-{\tan }^2\frac{x}{2}}=\frac{-1}{2}$.
Or
$4\tan \left(\frac{x}{2}\right)=-1+{\tan }^2\frac{x}{2}$.
${\tan }^2\frac{x}{2}-4\tan \left(\frac{x}{2}\right)-1=0$
Or
$\tan \left(\frac{x}{2}\right)=\frac{4\pm \sqrt{16+4}}{2}$

$=2\pm \sqrt{5}$