Solve $\sin 2 x - \cos 2 x = \sin x - \cos x$

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Detailed Solution

Given $\sin 2 x - \cos 2 x = \sin x - \cos x$

$\sin 2 x - \sin x = \cos 2 x - \cos x$

$\Rightarrow 2 \cos (\frac{2 x + x}{2}) \sin (\frac{2 x - x}{2})$ $= - 2 \sin (\frac{2 x + x}{2}) \sin (\frac{2 x - x}{2})$

$\Rightarrow \cos (\frac{3 x}{2}) \sin (\frac{x}{2}) + \sin (\frac{3 x}{2}) \sin (\frac{x}{2}) = 0$

$\Rightarrow \sin \frac{x}{2} [\cos \frac{3 x}{2} + \sin \frac{3 x}{2}] = 0$

$\sin \frac{x}{2} = 0$ or $\cos \frac{3 x}{2} + \sin \frac{3 x}{2} = 0$

G.S is $\frac{x}{2} = n π$ $\sin \frac{3 x}{2} = - \cos \frac{3 x}{2}$

$x = 2 n π, n \in Z$ $\tan \frac{3 x}{2} = - 1 = \tan \frac{3 π}{4}$

G.S is $\frac{3 x}{2} = n π + \frac{3 π}{4}$

$x = \frac{2 n π}{3} + \frac{π}{2}, n \in z$

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