If $\displaystyle \left ( \sqrt{3}-1 \right )\sin \theta +\left ( \sqrt{3}+1 \right )\cos \theta=2,$ then for $\displaystyle n \in I,\theta=$

Let $\sqrt{3} + 1 = r \cos \alpha$ and $\sqrt{3} - 1 = r \sin \alpha$

then, $r = \sqrt{(\sqrt{3} + 1)^2 + (\sqrt{3} -1)^2}$

$r = \sqrt{3 + 1 + 2\sqrt{3} + 3 + 1- 2\sqrt{3}}$

$r = \sqrt{8}$

$r = 2\sqrt{2}$

Also, $\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha}$

$\tan \alpha = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}$

$\tan \alpha = \dfrac{1 - \dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{\sqrt{3}}}$

$\tan \alpha = \tan (\dfrac{\pi}{4} - \dfrac{\pi}{6})$

$\tan \alpha = \tan \dfrac{\pi}{12}$

Now, $\displaystyle \left ( \sqrt{3}-1 \right )\sin \theta +\left ( \sqrt{3}+1 \right )\cos \theta=2$

$\displaystyle r\sin \alpha\sin \theta + r\cos \alpha\cos \theta=2$

$r\cos(\theta - \alpha) = 2$

$2 \sqrt{2} \cos(\theta - \dfrac{\pi}{12}) = 2$

$\cos (\theta - \dfrac{\pi}{12}) = \dfrac{1}{\sqrt{2}}$

$\theta - \dfrac{\pi}{12} = 2n\pi \pm \dfrac{\pi}{4}$

$\theta = 2n \pi \pm \dfrac{\pi}{4} + \dfrac{\pi}{12}$