Question

The general solution of equation $sinx+\sqrt{3}cosx=\sqrt{2}$ is

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

Answer 1

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

Given, $\sin x+\sqrt{3}\cos x=\sqrt{2}$
$\Rightarrow \frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x=\frac{1}{\sqrt{2}}$
$\Rightarrow \cos (x-\frac{\pi }{6})=\frac{1}{\sqrt{2}}$
$\Rightarrow x-\frac{\pi }{6}=2n\pi \pm \frac{\pi }{4}$

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