Solve for $x$ a trigonometric equation I want to solve for $x$ 
$$ {{2}^{{{\sin }^{4}}x-{{\cos }^{2}}x}}-{{2}^{{{\cos }^{4}}x-{{\sin }^{2}}x}}=\cos 2x $$
but I don't know how to start. Replacing $\sin x$ or $\cos x$ by $y$ led me nowhere because of the right side.
One of the solutions I've found is $x=\pi/4$ but there could be more solutions though.

Putting $\cos^2x=a,$
$\sin^4x-\cos^2x=(1-a)^2-a=a^2-3a+1$ and $\cos^4x-\sin^2x=a^2-(1-a)=a^2+a-1$
So we get, $2^{a^2-3a+1}-2^{a^2+a-1}=2a-1$
or, $2^{a^2-3a+1}(1-2^{4a-2})=2a-1$
If $2a-1>0,$  $4a-2>0\implies 2^{4a-2}>2^0=1\implies $ the left hand side is $<0$
Similarly, if $2a-1<0$  the right hand side is $>0$
If $2a-1=0,$  both sides become $0$
$\implies 2\cos^2x-1=0\iff \cos2x=0\implies 2x=(2n+1)\frac{\pi}2$ where $n$ is any integer 
