The general solution of sin x - cos x = sqrt{ | vedclass

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(B) Given equation: $\sin x - \cos x = \sqrt{2}$Divide both sides by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x = 1$Using the i

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$2n\pi + \frac{3\pi}{4}$

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(B) Given equation: $\sin x - \cos x = \sqrt{2}$Divide both sides by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x = 1$Using the identity $\sin(x - \frac{\pi}{4}) = \sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4}$:$\sin(x - \frac{\pi}{4}) = 1$We know that $\sin 	heta = 1$ implies $	heta = 2n\pi + \frac{\pi}{2}$ for any integer $n$.So,$x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{2}$$x = 2n\pi + \frac{\pi}{2} + \frac{\pi}{4}$$x = 2n\pi + \frac{3\pi}{4}$

The general solution of $\sin x - \cos x = \sqrt{2}$,for any integer $n$ is

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