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.
 A: 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 $\sin^2(x)=u=\frac{1}{2}$.
A: 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 $x=\pm\frac{\pi}{4}+2\pi\mathbb{Z},\pm\frac{3\pi}{4}+2\pi\mathbb{Z}$.
A: 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 
A: We study the injectivity function $f:[0, 1]\rightarrow R$, $f(t)= 2^{t^{2}+t-1}+t$
$f(t)= f(u)$ involving $t=u$ and then $sin^{2}x=cos^{2}x$...
