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.
 A: (Deleted comments refering to a previous version of the question which included a pair of seeming arithmetical errors).
You have the equation:
$$4^{\sin^2 x} = 4\pm 2\sqrt{3}.$$ Taking logarithms, you would get
$$\log(4^{\sin^2 x}) = \log(4\pm 2\sqrt{3}),$$
or equivalently,
$$\sin^2x \log(4) = \log(4\pm 2\sqrt{3}),$$
and hence
$$\sin^2 x = \frac{\log (4\pm 2\sqrt{3})}{\log 4}.$$
However, 
$$\frac{\log (4+ 2\sqrt{3})}{\log 4}\approx 1.44998,\qquad\text{and}\qquad \frac{\log(4 - 2\sqrt{3})}{\log 4}\approx -0.44998,$$
so neither quantity can equal $\sin^2 x$ with $x$ real (since $0\leq \sin^2 x \leq 1$ for all real numbers $x$). So there are no (real) solutions. 
Of course, Beni Bogossel's answer is better, but I write this in case you are interested in seeing how to carry your process all the way to the correct conclusion. 
A: 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 
   $$ t = \frac{1}{2} \  \ln  \left( {\frac {\ln  \left( 2 \right) }{\ln  \left( 2+\sqrt {3} \right) +\sqrt {\ln  \left( 4+2\,\sqrt {3} \right) \ln  \left( 1+1/
2\,\sqrt {3} \right) }}} \right) \approx -.6285882035
$$
A: This equality cannot happen. Since $\sin^2 x , \cos^2 x \in [0,1]$ it follows that $$4^{\sin^2 x}+4^{\cos^2 x}\leq 4+4=8$$ with equality when $\sin^2x=\cos^2x=1$. The last equality is impossible.
