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प्रश्न
Solve the following equation:
cosx + sin x = cos 2x + sin 2x
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उत्तर
\[\cos x + \sin x = \cos2x + \sin2x\]
\[ \Rightarrow \cos2x - \cos x + \sin2x - \sin x = 0\]
\[ \Rightarrow - 2\sin\frac{3x}{2}\sin\frac{x}{2} + 2\cos\frac{3x}{2}\sin\frac{x}{2} = 0\]
\[ \Rightarrow 2\sin\frac{x}{2}\left( \cos\frac{3x}{2} - \sin\frac{3x}{2} \right) = 0\]
\[ \Rightarrow 2 \sin\frac{x}{2} = 0\text{ or }\cos\frac{3x}{2} - \sin\frac{3x}{2} = 0\]
\[ \Rightarrow \sin\frac{x}{2} = 0\text{ or }\cos\frac{3x}{2} = \sin\frac{3x}{2}\]
\[ \Rightarrow \frac{x}{2} = n\pi\text{ or }\tan\frac{3x}{2} = 1\]
\[ \Rightarrow x = 2n\pi\text{ or }\tan\frac{3x}{2} = \tan\frac{\pi}{4}\]
\[ \Rightarrow x = 2n\pi\text{ or }\frac{3x}{2} = n\pi + \frac{\pi}{4}\]
\[ \Rightarrow x = 2n\pi\text{ or }3x = 2n\pi + \frac{\pi}{2}\]
\[ \Rightarrow x = 2n\pi\text{ or }x = \frac{2n\pi}{3} + \frac{\pi}{6}, n \in \mathbb{Z}\]
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