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 $2\sqrt{2}$ on both sides:$\frac{\sqrt{3}-1}{2\sqrt{2}} \sin 	heta

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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 $2\sqrt{2}$ on both sides:$\frac{\sqrt{3}-1}{2\sqrt{2}} \sin 	heta + \frac{\sqrt{3}+1}{2\sqrt{2}} \cos 	heta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$We know that $\cos(15^{\circ}) = \cos(\frac{\pi}{12}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin(15^{\circ}) = \sin(\frac{\pi}{12}) = \frac{\sqrt{3}-1}{2\sqrt{2}}$.Substituting these values,we get:$\sin(\frac{\pi}{12}) \sin 	heta + \cos(\frac{\pi}{12}) \cos 	heta = \frac{1}{\sqrt{2}}$Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$:$\cos(	heta - \frac{\pi}{12}) = \cos(\frac{\pi}{4})$The general solution for $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$.Therefore,$	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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