Question

The most general solution of $\sqrt{3}\cos \theta +\sin \theta =\sqrt{2}$ is.

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

Answer 1

Answer: (C)

$\cos 2\theta =\sin \alpha =\cos \left(\frac{\pi }{2}-\alpha \right)$
$\Rightarrow$ $2\theta$ = $2n\pi \pm \left(\frac{\pi }{2}-\alpha \right)$
$\Rightarrow$ $\theta$ = $n\pi \pm \left(\frac{\pi }{2}-\frac{\alpha }{2}\right)$