How to solve $2^{\sin^2(x)}=\cos(x)$

Question

$$2^{\sin^2(x)}=\cos(x)$$ 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?

Answer 1

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 $