Question

If $\sqrt{3}cos\Theta +sin\Theta =\sqrt{2},$ then the general solution of $\Theta$ is

Answer 1

$\sqrt{3}\cos \theta +\sin \theta =\sqrt{2}$

$\Rightarrow \frac{\sqrt{3}}{2}\cos \theta +\frac{\sin \theta }{2}=\frac{\sqrt{2}}{2}$

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

$\Rightarrow \sin \frac{\pi }{3}\cos \theta +\cos \frac{\pi }{3}\sin \theta =\sin \frac{\pi }{4}$

$\Rightarrow \sin (\frac{\pi }{3}+\theta )=\sin \frac{\pi }{4}$

$\Rightarrow \frac{\pi }{3}+\theta =n\pi +(-1{)}^n\frac{\pi }{4}$

$\Rightarrow \theta =n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{3}$.