Solve the following equations: 2^ ^2x +2 ^ 2x =22 - Vidyadip

Question

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

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Answer

$2^{\sin^2\text{x}}+2\cos^{2\text{x}}=2\sqrt{2}$ $\Rightarrow2^{\sin^2\text{x}}+2\cos^{-1\sin^2\text{x}}=2\sqrt{2}$ $\Rightarrow2^{\sin2\text{x}}+\frac{2}{2\sin^2\text{x}}2\sqrt{2}$ $\text{Let}2^{\sin^2\text{x}}+\frac{2}{2^{\sin2\text{x}}}=\text{y}$ $\Rightarrow\text{y}+\frac{2}{\text{y}}=2\sqrt{2}$ $\Rightarrow\text{y}^2+2=2\sqrt{2\text{y}}$ $\Rightarrow\text{y}^2-2\sqrt{2\text{y}}+2=0$ $\Rightarrow\text{y}^2-2\sqrt{2\text{y}}-\sqrt{2}\text{y}-\sqrt{2\text{y}}+2=0$ $\Rightarrow\text{y}\Big(\text{y}-\sqrt{2}\Big)-\sqrt{2}\Big(\text{y}-\sqrt{2}\Big)=0$ $\Rightarrow\Big(\text{y}-\sqrt{2}\Big)^2=0$ $\Rightarrow\Big(\text{y}-\sqrt{2}\Big)=0$ $\Rightarrow\text{y}=\sqrt{2}$ $\Rightarrow2^{\sin^2\text{x}}=2\frac{1}{2}$ $\Rightarrow\sin^2\text{x}=\frac{1}{2}$ $\Rightarrow\sin^2\text{x}=\sin^2\frac{\pi}{4}$ $\Rightarrow\text{x}=\text{n}\pi\pm\frac{\pi}{4},\text{n}\in\text{Z}$

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