Easiest way to solve $2^{\sin^2x}-2^{\cos^2x}=\cos 2x$

Question

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?

Answer 1

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 $$x=\dfrac{(2n+1)\pi}{4};\space\space n\in\mathbb Z$$

Answer 2

We have $2^{\sin^2x}+\sin^2x=2^{\cos^2x}+\cos^2x$, so consider the function $$f(x)=2^x+x$$ which is increasing and continuous then it is one-to-one and therefore $f(\sin^2x)=f(\cos^2x)$ gives us $\sin^2x=\cos^2x$.

Answer 3

Let $u:=\cos^2 x$ so $f(u):=2^{1-u}-2^u-2u+1=0=f(\frac12)$. But since $u\in[0,\,1]$, the AM-GM inequality implies $2^{1-u}+2^u\ge 2\sqrt{2}$ and $f^\prime(u)=(2^{1-u}+2^u)\ln 2-2\ge 2(\sqrt{2}-\ln 2)>0$, so $f$ is strictly increasing. The original equation is therefore equivalent to $\cos^2x=\frac12$.