Solving $\;2^{\large \cos x} = \sin x$ $$2^{\large \cos x} = |\sin x|$$
Solve the equation. I found just one solution $\cos x= 0$ and are there any other solutions. Right hand side is modulus $\sin x$.
 A: Graphs of $f(x)=|\sin x|$ and $f(x)=2^{\cos(x)}$. For the point $A\ $ $\cos x=0.56424...$

A: Letting $u=\cos(x)$, then we need to have
$$
4^u=1-u^2
$$
which implies
$$
u^2+4^u=1
$$
that is, $(u,2^u)$ crosses the unit circle.
$\hspace{4.5cm}$
Now that the equation is $2^{\large \cos(x)}=|\!\sin(x)|$, we also need to consider the curve $(u,-2^u)$, in red.
The point $u=0$ is there, as is the point $u=-0.82560777817003350220$. Thus, we have solutions
$$
\color{#00A000}{\cos^{-1}(0)=\frac\pi2}
$$
and
$$
\color{#C00000}{-\cos^{-1}(0)=-\frac\pi2}
$$
and
$$
\color{#00A000}{\cos^{-1}(-0.82560777817003350220)=2.5420748334255680556}
$$
and
$$
\color{#C00000}{-\cos^{-1}(-0.82560777817003350220)=-2.5420748334255680556}
$$
A: It would be enough to find the intersection of two graphs $2^{\cos x}$ and $\sin x$ over interval $[0,2\pi ]$ and since the graph again repeats after this the other solutions cvan be found by just adding $2k\pi$ where $k=0,1,2...$ to the intersection points. Hence plotting the graph once will help you a lot.
