Question

The general values of $\theta$ for which
$5\sin \theta -2{\cos }^2\theta -1=0$ is :

Answer 1

$5\sin \theta -2{\cos }^2\theta -1=0$

$\Rightarrow 2{\cos }^2\theta -5\sin \theta +1=0$

$\Rightarrow 2(1-{\sin }^2\theta )-5\sin \theta +1=0$

$\Rightarrow 2-2{\sin }^2\theta -5\sin \theta +1=0$

$\Rightarrow -2{\sin }^2\theta -5\sin \theta +3=0$

$\Rightarrow 2{\sin }^2\theta +5\sin \theta -3=0$

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

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

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

$\Rightarrow \sin \theta \ne -3,\sin \theta =\frac{1}{2}$

$\Rightarrow \theta =n\pi +{(-1)}^n\frac{\pi }{6}$