Question

Solve $\sqrt{3}cosx+sinx=\sqrt{2}.$

Answer 1

We have, $\sqrt{3}cosx+sinx=\sqrt{2}$

$\Rightarrow \frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx=\frac{\sqrt{2}}{2}$ [dividing both sides by 2]

$\Rightarrow cos\frac{\pi }{6}cosx+sin\frac{\pi }{6}sinx=\frac{1}{\sqrt{2}}$

$\Rightarrow cos(x-\frac{\pi }{6})=\frac{1}{\sqrt{2}}$ $[\because cos(A-B)=cosAcosB+sinAsinB]$

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

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

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

$\Rightarrow x=2n\pi +\frac{5\pi }{12}$ or $x=2n\pi -\frac{\pi }{12}$

Hence, $x=2n\pi +\frac{5\pi }{12},x=2n\pi -\frac{\pi }{12}$ where $n\in Z$