Question

Solve the following equation for $\theta$ being acute:
$2{\sin }^2\theta =3\cos \theta$.

Answer 1

$2(1-co{s}^2\theta )=3cos\theta$

$\Rightarrow 2co{s}^2\theta +3cos\theta -2=0$

$\Rightarrow 2co{s}^2\theta +4cos\theta -cos\theta -2=0$

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

$\Rightarrow cos\theta =\frac{1}{2}$ (since $cos\theta \ne -2)$

$\Rightarrow sin\theta =cos\frac{\pi }{3}$ So $\theta =\frac{\pi }{3}$