Question

General solution of the equation $(\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2$ is

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

Answer 1

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

The given equation is
$(\sqrt{3}-1)\sin \theta +(\sqrt{3}-1)\cos \theta =2$
Let $(\sqrt{3}-1)=r\sin \alpha ;(\sqrt{3}-1)=r\cos \alpha$
$\Rightarrow r=\sqrt{{(\sqrt{3}-1)}^2+{(\sqrt{3}+1)}^2}=2\sqrt{2}$
and $\tan \alpha =\frac{\sqrt{3}-1}{\sqrt{3}+1}=\tan {15}^o$
$\Rightarrow \alpha ={15}^o=\frac{\pi }{12}$
Therefore, $r\sin \alpha \sin \theta +r\cos \theta \cos \alpha =2$
$\Rightarrow 2\sqrt{2}\cos (\theta -\alpha )=2$
$\Rightarrow \theta -\frac{\pi }{12}=2n\pi \pm \frac{\pi }{4}$
$\Rightarrow \theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}$