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

Question

Solve the following equations:
$\cos\text{x}+\sin\text{x}=\cos2\text{x}+\sin2\text{x}$

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Answer

$\cos\text{x}+\sin\text{x}=\cos2\text{x}+\sin2\text{x}$$\Rightarrow\cos2\text{x}-\cos\text{x}+\sin2\text{x}-\sin\text{x}=0$
$\Rightarrow-2\sin\frac{3\text{x}}{2}\sin\frac{x}{2}+2\cos\frac{3\text{x}}{2}\sin\frac{\text{x}}{2}=0$
$\Rightarrow2\sin\frac{\pi}{\text{x}}\Big(\cos\frac{3\pi}{2}-\sin\frac{3\pi}{2}\Big)=0$
$\Rightarrow2\sin\frac{\text{x}}{2}=0$ or $\cos\frac{3\text{x}}{2}-\sin\frac{3\text{x}}{2}=0$
$\Rightarrow\sin\frac{\text{x}}{2}=0$ or $\cos\frac{3\text{x}}{2}=\sin\frac{3\text{x}}{2}$
$\Rightarrow\frac{\text{x}}{2}=\text{n}\pi$ or $\tan\frac{3\text{x}}{2}=1$
$\Rightarrow\text{x}=2\text{n}\pi$ or $\tan\frac{3\text{x}}{2}=\tan\frac{\pi}{4}$
$\Rightarrow\text{x}=\text{n}\pi$ or $\frac{3\pi}{2}-\text{n}\pi+\frac{\pi}{4}$
$\Rightarrow\text{x}=2\text{n}\pi$ or $3\text{x}=2\text{n}\pi+\frac{\pi}{2}$
$\Rightarrow\text{x}=2\text{n}\pi$ or $\text{x}=\frac{2\text{n}\pi}{3}+\frac{\pi}{6},\text{n}\in\text{Z}$

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