Question

The general solution of $(\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2$ is
(where $n\in Z$)

• A. $\theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}$
• B. $\theta =n\pi +(-1{)}^n\frac{\pi }{4}+\frac{\pi }{12}$
• C. $\theta =2n\pi \pm \frac{\pi }{4}$
• D. $\theta =n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{12}$

Answer 1

The correct option is A $\theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}$

$(\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2$
$\Rightarrow \frac{(\sqrt{3}-1)}{2\sqrt{2}}\sin \theta \left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)\cos \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \sin \frac{\pi }{12}\sin \theta +\cos \frac{\pi }{12}\cos \theta =\cos \frac{\pi }{4}$
$\Rightarrow \cos (\theta -\frac{\pi }{12})=\cos \frac{\pi }{4}$
$\Rightarrow \theta -\frac{\pi }{12}=2n\pi \pm \frac{\pi }{4},n\in Z$
So, $\theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12},n\in Z$