If sqrt{3} cos theta + sin theta = sqrt{2},th | vedclass

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Question

(D) Given equation: $\sqrt{3} \cos 	heta + \sin 	heta = \sqrt{2}$.Divide both sides by $\sqrt{(\sqrt{3})^2 + 1^2} = 2$:$\frac{\sqrt{3}}{2} \cos 	he

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$n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{3}$

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(D) Given equation: $\sqrt{3} \cos 	heta + \sin 	heta = \sqrt{2}$.Divide both sides by $\sqrt{(\sqrt{3})^2 + 1^2} = 2$:$\frac{\sqrt{3}}{2} \cos 	heta + \frac{1}{2} \sin 	heta = \frac{\sqrt{2}}{2}$.This can be written as $\sin \frac{\pi}{3} \cos 	heta + \cos \frac{\pi}{3} \sin 	heta = \frac{1}{\sqrt{2}}$.Using the formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we get:$\sin(	heta + \frac{\pi}{3}) = \sin \frac{\pi}{4}$.The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$.Therefore,$	heta + \frac{\pi}{3} = n\pi + (-1)^n \frac{\pi}{4}$.$	heta = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{3}$.

If $\sqrt{3} \cos \theta + \sin \theta = \sqrt{2}$,then the most general value of $\theta$ is

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