Question

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

• A. $2n\pi \pm \frac{\pi }{12}$
• B. $n\pi +(-1{)}^n\frac{\pi }{4}+\frac{\pi }{12}$
• C. $2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}$
• D. $n\pi +(-1{)}^n\pi +\frac{5\pi }{12}$

Answer 1

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

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, $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$

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

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

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

$\tan \alpha =\tan \frac{\pi }{12}$
Now, $(\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2$

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

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

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

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

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

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