Q. 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) $\theta = 2n\pi \pm \frac{\pi}{4} - \frac{\pi}{6}$
Explanation
(d)
Given equation can be rewritten as
$\frac{\sqrt{3}}{2}\cos{\theta + \frac{1}{2}\sin{\theta = \frac{\sqrt{2}}{2}}}$
$\Rightarrow \ \cos{\left( \theta + \frac{\pi}{6} \right) = \frac{1}{\sqrt{2}} = \cos\frac{\pi}{4}}$
$\Rightarrow \ \ \theta = 2n\pi \pm \frac{\pi}{4} - \frac{\pi}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.