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