Question

The general solution of $\sqrt{3}cosx+\sin x=\sqrt{2}$, for any integer $n$ is :

• A. $n\pi +\frac{\pi }{6}\pm \frac{\pi }{4}$
• B. $n\pi -\frac{\pi }{6}\pm \frac{\pi }{4}$
• C. $2n\pi -\frac{\pi }{6}\pm \frac{\pi }{4}$
• D. $2n\pi +\frac{\pi }{6}\pm \frac{\pi }{4}$

Answer 1

The correct option is D $2n\pi +\frac{\pi }{6}\pm \frac{\pi }{4}$

$\sqrt{3}cosx+\sin x=\sqrt{2}$

Dividing both sides with $\sqrt{3+1}=2$

$\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos (x-\frac{\pi }{6})=\frac{1}{\sqrt{2}}=\cos \frac{\pi }{4}$

$\Rightarrow x-\frac{\pi }{6}=2n\pi \pm \frac{\pi }{4}$

$\Rightarrow x=2n\pi +\frac{\pi }{6}\pm \frac{\pi }{4}$