Easiest way to solve $2^{\sin^2x}-2^{\cos^2x}=\cos 2x$ I'm trying to find best shortcut to crack this task 

$$2^{\sin^2x}-2^{\cos^2x}=\cos 2x$$

I tried first to go through using trigonometric identities 
$$\cos^2x+\sin^2x=1 \quad\text{and}\quad 2\sin^2x=1-\cos 2x$$
After substitution I reached this equation
$$2^{1-\cos x}-2^{(1-\cos 2x)/2}\cos 2x -2=0$$
Later, I supposed to use substitution to obtain,
$$u^4-2u^2-2u=0$$
where $u=\sqrt{v}$, $v=\frac{2}{w}$, and $w=\cos 2x$
As a result , I obtained two roots in $\mathbb{R}$ one of them is zero and the other is $1.76929$. This leads me to doubt in my process.
So, is there Hint or procedure can be followed to obtain the desire result?

HINT.-You have immediately a solution if you note that for  $x=\dfrac{\pi}{4}$ one has $\sin(x)=\cos(x)$ and $\cos(2x)=0$. The solution is done by $$x=\dfrac{(2n+1)\pi}{4};\space\space n\in\mathbb Z$$ 
