So that we have only one type of trig function in this, let's use the fact that

\(\displaystyle{{\cos}^{{2}}{\left({x}\right)}}+{{\sin}^{{2}}{\left({x}\right)}}={1}\rightarrow{{\cos}^{{{2}}}{x}}={1}-{{\sin}^{{{2}}}{x}}\)

to convert that cos^2 into an expression in sin^2.

\(\displaystyle{2}{{\cos}^{{{2}}}{x}}-{3}{\sin{{x}}}={0}\Leftrightarrow{2}-{2}{{\sin}^{{{2}}}{x}}-{3}{\sin{{x}}}={0}\Leftrightarrow{2}{{\sin}^{{{2}}}{x}}+{3}{\sin{{x}}}-{2}={0}\)

Now here's the trick: let y=sin(x). Then we have the quadratic equation

\(\displaystyle{2}{y}^{{{2}}}+{3}{y}−{2}={0}\)

Using the quadratic formula, we can easily see that the two solutions for this are

\(\displaystyle{y}=−{2},{\quad\text{or}\quad},{y}={\frac{{{1}}}{{{2}}}}\)

Now substitute y=sin(x) back in to get

\(\displaystyle{\sin{{\left({x}\right)}}}=−{2},{\quad\text{or}\quad},{\sin{{\left({x}\right)}}}={\frac{{{1}}}{{{2}}}}\)

The first of those has no solutions. The sine function is always between -1 and 1. So we just need the solutions to the second equation. Looking at our unit circle, we see that they are

\(\displaystyle{x}={30}^∘,{\quad\text{or}\quad},{x}={150}^∘\)