For n in Z,the general solution of the equati | 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{(3-2

Vedclass · Accepted Answer

$	heta = 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 $\sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} = \sqrt{(3-2\sqrt{3}+1) + (3+2\sqrt{3}+1)} = \sqrt{8} = 2\sqrt{2}$.$\frac{\sqrt{3}-1}{2\sqrt{2}} \sin 	heta + \frac{\sqrt{3}+1}{2\sqrt{2}} \cos 	heta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$Since $\cos \frac{\pi}{12} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin \frac{\pi}{12} = \frac{\sqrt{3}-1}{2\sqrt{2}}$,the equation becomes:$\sin \frac{\pi}{12} \sin 	heta + \cos \frac{\pi}{12} \cos 	heta = \cos \frac{\pi}{4}$$\cos(	heta - \frac{\pi}{12}) = \cos \frac{\pi}{4}$Using the general solution $\cos x = \cos \alpha \Rightarrow x = 2n\pi \pm \alpha$:$	heta - \frac{\pi}{12} = 2n\pi \pm \frac{\pi}{4}$$	heta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$,where $n \in Z$.

For $n \in Z$,the general solution of the equation $(\sqrt{3} - 1) \sin \theta + (\sqrt{3} + 1) \cos \theta = 2$ is

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