If sin theta + cos theta = 1,then the general | vedclass

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(B) Given $\sin 	heta + \cos 	heta = 1$. Dividing both sides by $\sqrt{1^2 + 1^2} = \sqrt{2}$,we get: $\frac{1}{\sqrt{2}} \sin 	heta + \frac{1}{\sq

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$n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}$

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(B) Given $\sin 	heta + \cos 	heta = 1$. Dividing both sides by $\sqrt{1^2 + 1^2} = \sqrt{2}$,we get: $\frac{1}{\sqrt{2}} \sin 	heta + \frac{1}{\sqrt{2}} \cos 	heta = \frac{1}{\sqrt{2}}$ $\sin 	heta \cos \frac{\pi}{4} + \cos 	heta \sin \frac{\pi}{4} = \sin \frac{\pi}{4}$ $\sin \left( 	heta + \frac{\pi}{4} ight) = \sin \frac{\pi}{4}$ The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$. Therefore,$	heta + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}$ $	heta = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}$.

If $\sin \theta + \cos \theta = 1$,then the general value of $\theta$ is

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