Question

If $2\sin {}^2\left(\right(\frac{\pi }{2})\cos {}^{2}x)=1-\cos \left(\pi \sin 2x\right);x\ne \frac{(2n+1)\pi }{2},n\in I$, then $\cos 2x$ is equal to

• A. $\frac{1}{5}$
• B. $\frac{3}{5}$
• C. $\frac{4}{5}$
• D. $1$

Answer 1

The correct option is B $\frac{3}{5}$

The given equation is equivalent to
$2si{n}^2\left(\right(\pi /2\left)co{s}^{2}x\right)=2si{n}^2\left(\right(\pi /2\left)sin2x\right)$
$\Rightarrow co{s}^{2}x=sin2x$
$\Rightarrow cosx(cosx-2sinx)=0$
$\Rightarrow 1-2tanx=0$ as $cosx\ne 0$,
$\because x\ne (2n+1)\pi /2$
$\Rightarrow tanx=1/2$
$\Rightarrow cos2x=\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}=\frac{3}{5}$