Question

Solve the following equation:
\[2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}\]

Answer 1

\[2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}\]
\[ \Rightarrow 2^{\sin^2 x} + 2^{1 - \sin^2 x} = 2\sqrt{2}\]
\[ \Rightarrow 2^{\sin^2 x} + \frac{2}{2^{\sin^2 x}} = 2\sqrt{2}\]
\[\text{ Let }2^{\sin^2 x} = y\]
\[ \Rightarrow y + \frac{2}{y} = 2\sqrt{2}\]
\[ \Rightarrow y^2 + 2 = 2\sqrt{2}y\]
\[ \Rightarrow y^2 - 2\sqrt{2}y + 2 = 0\]
\[ \Rightarrow y^2 - \sqrt{2}y - \sqrt{2}y + 2 = 0\]
\[ \Rightarrow y\left( y - \sqrt{2} \right) - \sqrt{2}\left( y - \sqrt{2} \right) = 0\]
\[ \Rightarrow \left( y - \sqrt{2} \right)^2 = 0\]
\[ \Rightarrow \left( y - \sqrt{2} \right) = 0\]
\[ \Rightarrow y = \sqrt{2}\]
\[ \Rightarrow 2^{\sin^2 x} = 2^\frac{1}{2} \]
\[ \Rightarrow \sin^2 x = \frac{1}{2}\]
\[ \Rightarrow \sin^2 x = \sin^2 \frac{\pi}{4}\]
\[ \Rightarrow x = n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z}\]