$4^{\sin^2x}+4^{\cos^2x}=8$
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.