<s>[INST] Find x in $4^{\sin^2x}+4^{\cos^2x}=8$ $$4^{\sin^2x}+4^{\cos^2x}=8$$
I solved like this:
\begin{align*}4^{\sin^2x}+4^{\cos^2x}=8&\Rightarrow4^{\sin^2x}+4^{1-\sin^2x}=8\\
&\Rightarrow4^{\sin^2x}+\frac{4}{4^{\sin^2x}}=8 |\cdot4^{\sin^2x}\\
&\Rightarrow4^{2\sin^2x}-8\cdot4^{\sin^2x}+4=0\\
y=4^{\sin^2x}&\Rightarrow y^2-8y+4=0\\
&\Rightarrow\Delta=64-16=48\\
&\Rightarrow y_{1,2}=\frac{8\pm 4\sqrt{3}}{2}\\
&\Rightarrow y_{1,2}=4\pm 2\sqrt{3}\\
&\Rightarrow 4^{\sin^2x}=4 \pm 2\sqrt{3}
\end{align*}
But now I'm stuck.
 [/INST] If you do want complex roots, one of the possible values of $x$ is $\arcsin(\sqrt{\log_4(4+2\sqrt{3})}) = \pi/2 + i t$
where 
   $$ t = \frac{1}{2} \  \ln  \left( {\frac {\ln  \left( 2 \right) }{\ln  \left( 2+\sqrt {3} \right) +\sqrt {\ln  \left( 4+2\,\sqrt {3} \right) \ln  \left( 1+1/
2\,\sqrt {3} \right) }}} \right) \approx -.6285882035
$$
</s>