The general solution of cos theta - sin theta | vedclass

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Question

(A) Divide the equation $\cos 	heta - \sin 	heta = 1$ by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \cos 	heta - \frac{1}{\sqrt{2}} \sin 	heta = \frac{1}{\sqr

Vedclass · Accepted Answer

$	heta = 2n\pi - \frac{\pi}{2}, 	heta = 2n\pi, n \in Z$

Vedclass · Answer

(A) Divide the equation $\cos 	heta - \sin 	heta = 1$ by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \cos 	heta - \frac{1}{\sqrt{2}} \sin 	heta = \frac{1}{\sqrt{2}}$$\cos 	heta \cos \frac{\pi}{4} - \sin 	heta \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$$\cos(	heta + \frac{\pi}{4}) = \cos \frac{\pi}{4}$The general solution for $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$:$	heta + \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}$Case $1$: $	heta + \frac{\pi}{4} = 2n\pi + \frac{\pi}{4} \implies 	heta = 2n\pi$Case $2$: $	heta + \frac{\pi}{4} = 2n\pi - \frac{\pi}{4} \implies 	heta = 2n\pi - \frac{\pi}{2}$Thus, $	heta = 2n\pi, 2n\pi - \frac{\pi}{2}$ where $n \in Z$.

The general solution of $\cos \theta - \sin \theta = 1$ is

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