Question

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

Answer 1

Answer: (A)

Let $\sqrt{3}+1=r\cos \alpha$ and $\sqrt{3}-1=r\sin \alpha$ $\therefore$ $r^2{\cos }^2\alpha +r^2{\sin }^2\alpha ={(\sqrt{3}+1)}^2+{(\sqrt{3}-1)}^2$
$\Rightarrow$ $r^2=3+1+2\sqrt{3}+3+1-2\sqrt{3}$
$\Rightarrow$ $r^2=8$
$\Rightarrow$ $r=2\sqrt{2}$ and $\tan \alpha =\frac{r\sin \alpha }{r\cos \alpha }=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$=\frac{1-1/\sqrt{3}}{1+1/\sqrt{3}}$
$=\frac{\tan \frac{\pi }{4}-\tan \frac{\pi }{6}}{1+\tan \frac{\pi }{4}.\tan \frac{\pi }{6}}$ $=\tan \left(\frac{\pi }{4}-\frac{\pi }{6}\right)$
$\Rightarrow$ $\tan \alpha =\tan (\pi /12)$ $\Rightarrow$ $\alpha =\frac{\pi }{12}$ Given, equation is $(\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2$
$\Rightarrow$ $r\sin \alpha \sin \theta +r\cos \alpha \cos \theta =2$
$\Rightarrow$ $2\sqrt{2}\cos (\theta -\alpha )=2$
$\Rightarrow$ $\cos \left(\theta -\frac{\pi }{12}\right)=\frac{2}{2\sqrt{2}}=\cos \left(\frac{\pi }{4}\right)$
$\Rightarrow$ $\theta -\frac{\pi }{12}=2n\pi \pm \frac{\pi }{4}$
$\Rightarrow$ $\theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}$