Question

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

• A. $n\pi +(-1{)}^n\frac{\pi }{4}$
• B. $(-1{)}^n\frac{\pi }{4}-\frac{\pi }{3}$
• C. $n\pi +\frac{\pi }{4}-\frac{\pi }{3}$
• D. $n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{3}$

Answer 1

The correct option is D $n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{3}$

$\sqrt{3}\cos \theta +\sin \theta =\sqrt{2}$
$\Rightarrow \frac{\sqrt{3}}{2}\cos \theta +\frac{1}{2}\sin \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \sin \left(\frac{\pi }{3}\right)\cos \theta +\cos \left(\frac{\pi }{3}\right)\sin \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \sin (\frac{\pi }{3}+\theta )=\frac{1}{\sqrt{2}}$
$\Rightarrow \frac{\pi }{3}+\theta =n\pi +{(-1)}^n\frac{\pi }{4}\Rightarrow \theta =n\pi +{(-1)}^n\frac{\pi }{4}-\frac{\pi }{3}$