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

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Question

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

Vedclass · Accepted Answer

$2 n \pi \pm \frac{\pi}{4} + \frac{\pi}{12}$

Vedclass · Answer

(A) Given equation: $(\sqrt{3}-1) \sin 	heta + (\sqrt{3}+1) \cos 	heta = 2$. Divide both sides by $\sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} = \sqrt{3+1-2\sqrt{3} + 3+1+2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$. $\frac{\sqrt{3}-1}{2\sqrt{2}} \sin 	heta + \frac{\sqrt{3}+1}{2\sqrt{2}} \cos 	heta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$. Let $\cos \alpha = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin \alpha = \frac{\sqrt{3}-1}{2\sqrt{2}}$. Then $	an \alpha = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 	an(60^\circ - 45^\circ) = 	an(15^\circ) = 	an(\frac{\pi}{12})$. So,$\alpha = \frac{\pi}{12}$. The equation becomes $\cos \alpha \cos 	heta + \sin \alpha \sin 	heta = \frac{1}{\sqrt{2}}$. $\cos(	heta - \alpha) = \cos(\frac{\pi}{4})$. The general solution is $	heta - \alpha = 2n\pi \pm \frac{\pi}{4}$. $	heta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$.

The general solution of the equation $(\sqrt{3}-1) \sin \theta + (\sqrt{3}+1) \cos \theta = 2$ is

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