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

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(A) Given,$\sin x - \cos x = \sqrt{2}$ Divide both sides by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x = 1$ Using $\cos \frac{

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

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(A) Given,$\sin x - \cos x = \sqrt{2}$ Divide both sides by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x = 1$ Using $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we get: $\sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} = 1$ Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $\sin(x - \frac{\pi}{4}) = 1$ Since $\sin 	heta = 1$ implies $	heta = 2n\pi + \frac{\pi}{2}$: $x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{2}$ $x = 2n\pi + \frac{\pi}{2} + \frac{\pi}{4}$ $x = 2n\pi + \frac{3\pi}{4}$

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

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