Question

Solve for $\theta$ if $4{\cos }^2\theta$ + 2$\sin \theta$ – 4 = 0, where $0^{\circ }$ $\le$ $\theta$ $\le$ ${90}^{\circ }$.

Answer 1

$\begin{array}{cc}4{\cos }^2\theta +2\sin \theta -4& =0\\ 4(1-{\sin }^2\theta )+2\sin \theta & =0\\ 4-4{\sin }^2\theta +2\sin \theta -4& =0\\ 2\sin \theta (1-2\sin \theta )& =0\end{array}$


$\sin \theta =0$ or $\sin \theta =\frac{1}{2}$
So,
$\theta =0^{\circ }$ or $\theta ={30}^{\circ }$