Question

Find the equation of the problem

Answer 1

$6\sin^2\theta-\sin\theta=1\ \ \ -1\\\\6\sin^2\theta-\sin\theta-1=0\\\\6\sin\theta-3\sin\theta+2\sin\theta-1=0\\\\3\sin\theta(2\sin\theta-1)+1(2\sin\theta-1)=0\\\\(2\sin\theta-1)(3\sin\theta+1)=0\iff2\sin\theta-1=0\ \vee\ 3\sin\theta+1=0\\\\2\sin\theta=1\ \ |:2\ \ \vee\ \ 3\sin\theta=-1\ \ |:3\\\\\sin\theta=\dfrac{1}{2}\ \vee\ \sin\theta=-\dfrac{1}{3}$
$\theta=30^o\ \vee\ \theta=150^o\ \vee\ \theta=360^o-\sin^{-1}\left(\dfrac{1}{3}\right)\ \vee\ \theta=180^o+\sin^{-1}\left(\dfrac{1}{3}\right)$