Question

If $\cos 2x+2\cos x=1$, then ${\sin }^{2}x(2-{\cos }^{2}x)$ is

• A. $1$
• B. $2$
• C. $4$
• D. none of these

Answer 1

Answer: (A)

$1$

Solution:- (A) $1$

$\cos 2x+2\cos x=1$

$2{\cos }^{2}x-1+2\cos x=1(\because \cos 2x=2{\cos }^{2}x-1)$

$\Rightarrow 2{\cos }^{2}x+2\cos x-2=0$

$\Rightarrow {\cos }^{2}x+\cos x=1.....\left(1\right)$

$\Rightarrow 1-{\cos }^{2}x=\cos x$

$\Rightarrow {\sin }^{2}x=\cos x.....\left(2\right)$

Now,

${\sin }^2(2-{\cos }^{2}x)$

$={\sin }^{2}x(1+{\sin }^{2}x)$

$=\cos x(1+\cos x)\left[\text{From}\left(2\right)\right]$

$=\cos x+{\cos }^{2}x$

$=1\left[\text{From}\left(1\right)\right]$

Hence ${\sin }^2(2-{\cos }^{2}x)=1$.