Question

If $\sqrt{3}\cos \theta +\sin \theta =\sqrt{2},$ then the most general value of $\theta$ is :

• 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 $n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{3}$

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

$\sin {60}^o\cos \theta +\cos {60}^o\sin \theta =\frac{1}{\sqrt{2}}$

$\sin (60+\theta )=\frac{1}{\sqrt{2}}$

$60+\theta =n\pi +(-1{)}^n\frac{\pi }{4}$

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

$\therefore \left(D\right)$ Answer.