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.

Solution 1:

Let $u=\sin^2(x)$ and $v=\cos^2(x)$, then $2^{u^2-v}-2^{v^2-u} = v-u$ and $u+v=1$. Thus $2^{u^2+u-1}+u = 2^{v^2+v-1}+v$.

Define $f(u) = 2^{u^2+u-1}+u$, then we are looking for a $u\in[0,1]$ such that $f(u)=f(1-u)$. However $f^\prime(u) = \ln(2)(2u+1)2^{u^2+u-1}+1>0$ for all $u\in[0,1]$. Hence $f$ is injective on $[0,1]$ and thus $f(u)=f(1-u)$ if and only if $u=1-u$. Thus $\sin^2(x)=u=\frac{1}{2}$.

Solution 2:

The left side of the equation is greater than $0$ if and only if the right side is less than $0$, and vice versa. This follows from $$ [\sin^4x-\cos^2x]-[\cos^4x-\sin^2x]=-\left(\cos^2x-\sin^2x\right)(\cos^2x+\sin^2x+1)=-2\cos2x.$$ Therefore all possible solutions correspond to zeros of the right hand side (which are also automatically zeroes of the left hand side). This gives $x=\pm\frac{\pi}{4}+2\pi\mathbb{Z},\pm\frac{3\pi}{4}+2\pi\mathbb{Z}$.