Question

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

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

Answer 1

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

Let $\sqrt{3}+1=r\cos \alpha$, and $\sqrt{3}-1=2\sin \alpha$

$\therefore r^2={(\sqrt{3}+1)}^2+{(\sqrt{3}-1)}^2=8$

and $\tan \alpha =\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\tan (\frac{\pi }{4}-\frac{\pi }{6})$

$=\tan \left(\frac{\pi }{12}\right)\Rightarrow \alpha =\frac{\pi }{12}$

From the given equation, we get
$r\cos (\theta -\alpha )=2\Rightarrow \cos (\theta -\frac{\pi }{12})=\frac{1}{\sqrt{2}}=\cos \left(\frac{\pi }{4}\right)$

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