If sin theta + cos theta = sqrt{2} cos alpha, | vedclass

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(B) Given: $\sin 	heta + \cos 	heta = \sqrt{2} \cos \alpha$Divide both sides by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \sin 	heta + \frac{1}{\sqrt{2}} \cos

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$2n\pi + \frac{\pi}{4} \pm \alpha$

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(B) Given: $\sin 	heta + \cos 	heta = \sqrt{2} \cos \alpha$Divide both sides by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \sin 	heta + \frac{1}{\sqrt{2}} \cos 	heta = \cos \alpha$Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$:$\cos 	heta \cos \frac{\pi}{4} + \sin 	heta \sin \frac{\pi}{4} = \cos \alpha$$\cos \left( 	heta - \frac{\pi}{4} ight) = \cos \alpha$The general solution for $\cos x = \cos y$ is $x = 2n\pi \pm y$.Therefore,$	heta - \frac{\pi}{4} = 2n\pi \pm \alpha$$	heta = 2n\pi + \frac{\pi}{4} \pm \alpha$.

If $\sin \theta + \cos \theta = \sqrt{2} \cos \alpha$,then the general value of $\theta$ is

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