Question

Solve $si{n}^2\theta -cos\theta =\frac{1}{4}$ for $\theta$ & $0\le \theta \le 2\pi$

Answer 1

${\sin }^2\theta -\cos \theta =\frac{1}{4}$

$\Rightarrow 4{\sin }^2\theta -4\cos \theta =1$

$4-4{\cos }^2\theta -4\cos \theta =1$

$4{\cos }^2\theta +4\cos \theta +1-4=0$

$4{\cos }^2\theta +4\cos \theta -3=0$

$\Rightarrow 4{\cos }^2\theta +6\cos \theta -2\cos \theta -3=0$

$2\cos theta(2\cos \theta +3)-(2\cos \theta +3)=0$

$\Rightarrow (2\cos \theta -1)(2\cos \theta +3)=0$

$2\cos \theta -1=0$

$\cos \theta =\frac{1}{2}$

$\cos \theta =\cos \frac{\pi }{3}$

$\Rightarrow \theta =\frac{\pi }{3}$

Hence, solved.