From Förberedande kurs i matematik 1

If we use the Pythagorean identity and write \displaystyle \sin^2\!x as \displaystyle 1-\cos^2\!x, the whole equation can be written in terms of \displaystyle \cos x,

\displaystyle 2(1-\cos^2\!x) - 3\cos x = 0\,,

or, in rearranged form,

\displaystyle 2\cos^2\!x + 3\cos x - 2 = 0\,\textrm{.}

With the equation expressed entirely in terms of \displaystyle \cos x, we can introduce a new unknown variable \displaystyle t=\cos x and solve the equation with respect to t. Expressed in terms of t, the equation is

\displaystyle 2t^2+3t-2 = 0

and this quadratic equation has the solutions \displaystyle t=\tfrac{1}{2} and \displaystyle t=-2\,.

In terms of x, this means that either \displaystyle \cos x = \tfrac{1}{2} or \displaystyle \cos x = -2. The first case occurs when

\displaystyle x=\pm \frac{\pi}{3}+2n\pi\qquad(n is an arbitrary integer),

whilst the equation \displaystyle \cos x = -2 has no solutions at all (the values of cosine lie between -1 and 1).

The answer is that the equation has the solutions

\displaystyle x = \pm\frac{\pi}{3} + 2n\pi\,,

where n is an arbitrary integer.