Solve the following equation sin x + sqrt{3} | vedclass

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Question

(A) Given equation: $\sin x + \sqrt{3} \cos x = \sqrt{2}$.Divide both sides by $2$:$\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x = \frac{\sqrt{2}}{2

Vedclass · Accepted Answer

$x = 2n\pi + \frac{\pi}{6} \pm \frac{\pi}{4}$

Vedclass · Answer

(A) Given equation: $\sin x + \sqrt{3} \cos x = \sqrt{2}$.Divide both sides by $2$:$\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x = \frac{\sqrt{2}}{2}$.This can be written as:$\sin x \cos(\frac{\pi}{3}) + \cos x \sin(\frac{\pi}{3}) = \frac{1}{\sqrt{2}}$.Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$:$\sin(x + \frac{\pi}{3}) = \sin(\frac{\pi}{4})$.The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n\pi + (-1)^n \alpha$.However,using the form $\cos(x - \alpha) = \cos \beta$ is often simpler for this type:$\sin x + \sqrt{3} \cos x = 2(\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x) = 2 \cos(x - \frac{\pi}{6}) = \sqrt{2}$.$\cos(x - \frac{\pi}{6}) = \frac{\sqrt{2}}{2} = \cos(\frac{\pi}{4})$.Therefore,$x - \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{4}$.$x = 2n\pi + \frac{\pi}{6} \pm \frac{\pi}{4}$.

Solve the following equation $\sin x + \sqrt{3} \cos x = \sqrt{2}$.

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