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

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(A) Divide the equation by $r = \sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} = \sqrt{3-2\sqrt{3}+1 + 3+2\sqrt{3}+1} = \sqrt{8} = 2\sqrt{2}$.The equation bec

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

Vedclass · Answer

(A) Divide the equation by $r = \sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} = \sqrt{3-2\sqrt{3}+1 + 3+2\sqrt{3}+1} = \sqrt{8} = 2\sqrt{2}$.The equation becomes $\frac{\sqrt{3}-1}{2\sqrt{2}}\sin 	heta + \frac{\sqrt{3}+1}{2\sqrt{2}}\cos 	heta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.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 is $\cos \alpha \cos 	heta + \sin \alpha \sin 	heta = \cos(\frac{\pi}{4})$.$\cos(	heta - \alpha) = \cos(\frac{\pi}{4})$.$	heta - \frac{\pi}{12} = 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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