If n is any integer,then the general solution | vedclass

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(C) Given equation is $\cos x - \sin x = \frac{1}{\sqrt{2}}$.Dividing both sides by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\s

Vedclass · Accepted Answer

$x = 2n\pi + \frac{\pi}{12}$ or $x = 2n\pi - \frac{7\pi}{12}$

Vedclass · Answer

(C) Given equation is $\cos x - \sin x = \frac{1}{\sqrt{2}}$.Dividing both sides by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x = \frac{1}{2}$.This can be written as $\cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4} = \frac{1}{2}$.Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we get $\cos(x + \frac{\pi}{4}) = \frac{1}{2}$.Since $\cos \frac{\pi}{3} = \frac{1}{2}$,the equation becomes $\cos(x + \frac{\pi}{4}) = \cos \frac{\pi}{3}$.The general solution for $\cos 	heta = \cos \alpha$ is $	heta = 2n\pi \pm \alpha$.Therefore,$x + \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{3}$.Case $1$: $x = 2n\pi + \frac{\pi}{3} - \frac{\pi}{4} = 2n\pi + \frac{\pi}{12}$.Case $2$: $x = 2n\pi - \frac{\pi}{3} - \frac{\pi}{4} = 2n\pi - \frac{7\pi}{12}$.

If $n$ is any integer,then the general solution of the equation $\cos x - \sin x = \frac{1}{\sqrt{2}}$ is

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