The general solution of the equation sqrt{3} | vedclass

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Question

(C) Given equation: $\sqrt{3} \cos 	heta + \sin 	heta = \sqrt{2}$ Divide both sides by $2$: $\frac{\sqrt{3}}{2} \cos 	heta + \frac{1}{2} \sin 	het

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given equation: $\sqrt{3} \cos 	heta + \sin 	heta = \sqrt{2}$ Divide both sides by $2$: $\frac{\sqrt{3}}{2} \cos 	heta + \frac{1}{2} \sin 	heta = \frac{\sqrt{2}}{2}$ Using $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{3} = \frac{1}{2}$,we get: $\sin \frac{\pi}{3} \cos 	heta + \cos \frac{\pi}{3} \sin 	heta = \frac{1}{\sqrt{2}}$ Applying the formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$: $\sin \left(	heta + \frac{\pi}{3}ight) = \sin \frac{\pi}{4}$ The general solution for $\sin x = \sin \alpha$ is $x = n \pi + (-1)^{n} \alpha$: $	heta + \frac{\pi}{3} = n \pi + (-1)^{n} \frac{\pi}{4}, n \in Z$ Therefore,$	heta = n \pi + (-1)^{n} \frac{\pi}{4} - \frac{\pi}{3}, n \in Z$

The general solution of the equation $\sqrt{3} \cos \theta + \sin \theta = \sqrt{2}$ is

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