Question

Solve ${\sin }^2\theta -\cos \theta =\frac{1}{4}.0\le \theta \le 2\pi$

Answer 1

Given, ${\sin }^2\theta -\cos \theta =\frac{1}{4}$
$\Rightarrow {\cos }^2\theta -\cos \theta =\frac{3}{4}$
$\Rightarrow \cos \theta =\frac{-1\pm \sqrt{4}}{2}=-\frac{3}{2},\frac{1}{2}$
$\cos \theta \ne -\frac{3}{2}$ or $\cos \theta =\frac{1}{2}$
$\Rightarrow \theta =2n\pi \pm \frac{\pi }{3}$