The set of solutions of the equation (sqrt{3} | vedclass

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

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

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(A) Given equation: $(\sqrt{3}-1) \sin 	heta+(\sqrt{3}+1) \cos 	heta=2$Divide by $\sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} = \sqrt{3+1-2\sqrt{3} + 3+1+2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$.$\frac{\sqrt{3}-1}{2\sqrt{2}} \sin 	heta + \frac{\sqrt{3}+1}{2\sqrt{2}} \cos 	heta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.We know $\cos 15^{\circ} = \cos \frac{\pi}{12} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin 15^{\circ} = \sin \frac{\pi}{12} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.So,$\sin \frac{\pi}{12} \sin 	heta + \cos \frac{\pi}{12} \cos 	heta = \cos \frac{\pi}{4}$.$\cos(	heta - \frac{\pi}{12}) = \cos \frac{\pi}{4}$.General solution: $	heta - \frac{\pi}{12} = 2n\pi \pm \frac{\pi}{4}$.$	heta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$,where $n \in Z$.

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

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