Question

Solve the following equations:
$2^{{\sin }^{2}x}+2^{{\cos }^{2}x}=2\sqrt{2}$

Answer 1

$$\begin{gathered} 2^{{\sin }^{2}x}+2^{{\cos }^{2}x}=2\sqrt{2} \\ \Rightarrow 2^{{\sin }^{2}x}+2^{1-{\sin }^{2}x}=2\sqrt{2} \\ \Rightarrow 2^{{\sin }^{2}x}+\frac{2}{2^{{\sin }^{2}x}}=2\sqrt{2} \\ \mathrm{Let}{2}^{{\sin }^{2}x}=y \\ \Rightarrow y+\frac{2}{y}=2\sqrt{2} \\ \Rightarrow y^2+2=2\sqrt{2}y \\ \Rightarrow y^2-2\sqrt{2}y+2=0 \\ \Rightarrow y^2-\sqrt{2}y-\sqrt{2}y+2=0 \\ \Rightarrow y\left(y-\sqrt{2}\right)-\sqrt{2}\left(y-\sqrt{2}\right)=0 \\ \Rightarrow {\left(y-\sqrt{2}\right)}^2=0 \\ \Rightarrow \left(y-\sqrt{2}\right)=0 \\ \Rightarrow y=\sqrt{2} \\ \Rightarrow 2^{{\sin }^{2}x}=2^{\frac{1}{2}} \\ \Rightarrow {\sin }^{2}x=\frac{1}{2} \\ \Rightarrow {\sin }^{2}x={\sin }^2\frac{\mathrm{\pi }}{4} \\ \Rightarrow x=n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4},n\in \mathrm{\mathbb{Z}} \end{gathered}$$