Question

If $2 \,sin^2 ((\pi/2) cos^2x) = 1 - cos(\pi\, sin \,2x), x \ne (2n + 1) \pi/2, n \in I$, then $cos\,2x$ is equal to _____ .

Answer 1

The given equation is equivalent to
$2 \,sin^2 ((\pi/2) cos^2x ) = 2\, sin^2 ((\pi/2) sin\,2x)$
or $cos^2x = sin\,2x$
or $cos\,x (cos\,x - 2 \,sin\,x) = 0$
$\Rightarrow 1 - 2\,tan\,x = 0$ as $cos\,x \ne 0, x \ne (2n + 1) \frac{\pi}{2}$
or $tan\,x = \frac{1}{2}$
$\Rightarrow cos\,2x = \frac{1 - tan^2\,x}{1 + tan^2\,x} = \frac{3}{5}$