How to solve $2^{\sin^2(x)}=\cos(x)$ $$2^{\sin^2(x)}=\cos(x)$$
How can I find the solution?
This is what I did :
$$\ln(2^{\sin^2(x)})=\ln(\cos(x))$$
$$\sin^2(x)\ln(2)=\ln(\cos(x))$$
$$(1-\cos^2(x))\ln(2)=\ln(\cos(x))$$
$$\cos^2(x)\ln(2)+\ln(\cos(x))-\ln(2)=0$$
I put : $t=\cos(x)$
$$t^2\ln(2)+\ln(t)-\ln(2)=0$$
If I want to take out the $t$ from the $\ln$, I will have more complicated equation with $\exp$ function.
I don't know if what I did helps to get the result. Can you help me please?

Range of $2^{(\sin x)^2}$ is $[1,2]$ whereas range of $\cos x$ is $[-1,1]$, hence the only solutions is $\cos x = 2^{(\sin x)^2} = 1$
$\cos x = 1$ at $2n \pi $
$2^{(\sin x)^2} = 1 $
$\implies (\sin x) = 0$
$\implies x = n \pi $
The solutions have an intersection,
$x = 2n\pi $
