The general solution of the trigonometric equ | 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

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

Vedclass · Answer

(A) Given equation: $(\sqrt{3}-1) \sin 	heta + (\sqrt{3}+1) \cos 	heta = 2$. Divide both sides by $\sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} = \sqrt{(3-2\sqrt{3}+1) + (3+2\sqrt{3}+1)} = \sqrt{8} = 2\sqrt{2}$. $\frac{\sqrt{3}-1}{2\sqrt{2}} \sin 	heta + \frac{\sqrt{3}+1}{2\sqrt{2}} \cos 	heta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$. 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(75^\circ) = 	an(\frac{5\pi}{12})$. So,$\sin(	heta + \alpha) = \sin(\frac{\pi}{4})$. Thus,$	heta + \frac{5\pi}{12} = n\pi + (-1)^n \frac{\pi}{4}$. $	heta = n\pi + (-1)^n \frac{\pi}{4} - \frac{5\pi}{12}$. Alternatively,using $\cos(	heta - \beta) = \frac{1}{\sqrt{2}}$ where $\beta = \frac{\pi}{12}$,we get $	heta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$.

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

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