Solve for $x$ where $\sin(x)^{\cos(x)}$= $2$ I found this question in a mathematics book (some kind of challenge questions book). I was just trying to solve this question for fun. At the moment I am a bit confused. So by now, I have tried two things 1st of all is to sub $cos(x)$ as
$\frac{1}{\pm \sqrt{1+\tan^2(x)}}$
and
$\pm \sqrt{\sin^2(x)-1}$.
However, I haven't made any progress. I graphed both equation son desmos and Graph of both sides of the equation. I also got this solution from a math computation website solution.
Any help will be highly appreciated.

First approach
Plotting or by inspection, the first solution is close to $\frac {5  \pi}6$.
Now, consider
$$f(x)=\cos(x)\log(\sin(x))-\log(2)$$ and expand as a Taylor series
$$f(x)=\frac{1}{2} \left(\sqrt{3}-2\right) \log (2)+\frac{1}{2} \left(x-\frac{5 \pi
   }{6}\right) (3+\log (2))-$$ $$\frac{1}{4} \left(x-\frac{5 \pi }{6}\right)^2
   \left(\sqrt{3} (\log (2)-6)\right)+\left(x-\frac{5 \pi }{6}\right)^3
   \left(\frac{9}{4}-\frac{\log (2)}{12}\right)+$$ $$O\left(\left(x-\frac{5 \pi
   }{6}\right)^4\right)$$
Using series reversion
$$x=\frac{5 \pi }{6}+t+\frac{\sqrt{3}  (\log (2)-6)}{6+\log (4)}t^2+\frac{ (243+\log
   (4) (\log (32)-66))}{6 (3+\log (2))^2}t^3+O\left(t^{4}\right)$$ where $t=\frac{2 f(x)+(2-\sqrt{3}) \log (2)}{3+\log (2)}$.
Since you want $f(x)=0$, then the approximation
$x \approx 2.6653795$ while the "exact" solution, given by Newton method, is
$x=2.6653571$
Second approach
Using the tangent half-angle substitution $x=2\tan^{-1}(t)$, you need to find the zero of function
$$g(t)=\left(\frac{2 t}{t^2+1}\right)^{\frac{1-t^2}{1+t^2}}-2$$ Plotting, the solution is close to $t=4$ and the function is very close to linearity. Using Newton method, the iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 4.000000000 \\
 1 & 4.120370127 \\
 2 & 4.119928024 \\
 3 & 4.119928018
\end{array}
\right)$$ Back to $x$, $x=2.6653571$.
Third approach
The derivative of $f(x)$ cancels at $x=\frac \pi 2$ and the second derivative test shows that it is a local minimum. Expand as a series
$$f(x)=\frac{1}{2} \left(x-\frac{\pi }{2}\right)^3+\frac{1}{80} \left(x-\frac{\pi
   }{2}\right)^7+O\left(\left(x-\frac{\pi }{2}\right)^9\right)-\log(2)$$ Using the first term only, a first estimate
$$x_0=\frac{\pi }{2}+\sqrt[3]{2 \log (2)}=2.68582$$ Plug this is the second term
$$0=\frac{1}{2} \left(x-\frac{\pi }{2}\right)^3+\frac{1}{80} \log ^{\frac{7}{3}}(4)-\log(2)$$
$$x_1=\frac{\pi }{2}+\sqrt[3]{\log (4)-\frac{1}{40} \log ^{\frac{7}{3}}(4)}=2.67127$$
