# Find the general solution of the equation $(\sqrt 3+1)\cos \theta+(\sqrt 3-1)\sin \theta=2$

$\begin{array}{1 1}(A)\;2n\pi\pm \large\frac{\pi}{4}+\frac{\pi}{12}&(B)\;n\pi\pm \large\frac{\pi}{6}\\(C)\;n\pi\pm \large\frac{\pi}{8}&(D)\;2n\pi\pm \large\frac{3\pi}{4}\end{array}$

Toolbox:
• $a\cos \theta+b\sin \theta=c$
Given : $(\sqrt 3+1)\cos\theta +(\sqrt 3-1)\sin \theta=2$
Let $(\sqrt 3+1)=r\cos \alpha$
$(\sqrt 3-1)=r\sin \alpha$
$r\cos \alpha\cos \theta+r\sin \alpha\sin \theta=2$
$r(\cos \theta\cos \alpha+\sin \theta \sin \alpha)=2$
$r(\cos(\theta-\alpha)=2$
$\cos (\theta-\alpha)=\large\frac{2}{r}$
$\cos (\theta-\alpha)=\large\frac{1}{\sqrt 2}$$=\cos \large\frac{\pi}{4} r=\sqrt{(\sqrt 3-1)^2+(\sqrt 3+1)^2} \Rightarrow \sqrt{4-2\sqrt 3+4+2\sqrt 3} \Rightarrow \sqrt 8=2\sqrt 2 \tan \alpha=\large\frac{\sqrt 3-1}{\sqrt 3+1}=$$\tan (\large\frac{\pi}{3}-\frac{\pi}{4})$
$\alpha=\large\frac{\pi}{12}$
$(\theta-\alpha)=2n\pi\pm \large\frac{\pi}{4}$
$\alpha=\large\frac{\pi}{12}$
$\theta=2n\pi\pm \large\frac{\pi}{4}+\frac{\pi}{12}$
Hence (A) is the correct answer.