The set of solutions of the equation ( 3 -1) θ+( 3 +1) θ=2 is

$(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2\frac{\sqrt{3}-1}{2} \sin \theta+\frac{\sqrt{3}+1}{2} \cos \theta=1$ ...(i) Comparing with $a \sin \theta+b \cos \theta=1$. ie, $\quad a=\frac{\sqrt{3}-1}{2}, b=\frac{\sqrt{3}+1}{2}\sqrt{a^2+b^2}=\sqrt{\frac{(\sqrt{3}-1)^2}{4}+\frac{(\sqrt{3}+1)^2}{4}}=\frac{1}{2} \sqrt{3+1-2 \sqrt{3}+3+1+2 \sqrt{3}}=\frac{1}{2} \sqrt{8}=\frac{1}{2} \cdot 2 \sqrt{2}=\sqrt{2}$ Dividing on both sides by $\sqrt{2}$ in Eq. (i), we get \left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right) \sin \theta+\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right) \cos \theta=\frac{1}{\sqrt{2}}$Let$\quad \sin \alpha=\frac{\sqrt{3}-1}{2 \sqrt{2}}$Then,$\cos \alpha=\sqrt{1-\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^2}=\sqrt{1-\frac{(4-2 \sqrt{3})}{8}}=\sqrt{\frac{8-4+2 \sqrt{3}}{8}}=\sqrt{\frac{4+2 \sqrt{3}}{8}}=\sqrt{\frac{\sqrt{3}+1}{4}}=\sqrt{\frac{(\sqrt{3}+1)^2}{8}}= $\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right) So,$\quad \sin \alpha \cdot \sin \theta+\cos \alpha \cdot \cos \theta=\frac{1}{\sqrt{2}}\cos (\theta-\alpha)=\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}\theta-\alpha=$2 n \pi \pm$\frac{\pi}{4}, \theta=$2 n \pi \pm$\frac{\pi}{4}+\alpha\ldots (ii) $\because \quad \cos$15=\frac{\sqrt{3}+1}{2 \sqrt{2}} ie, $\quad \cos \alpha=\cos 15^{\circ}=\cos$\frac{\pi}{12}\Rightarrow \quad \alpha=\frac{\pi}{12} From Eq. (ii), \left[\theta=2 n \pi \pm \frac{\pi}{4}+\frac{\pi}{12}\right] .n \in Z