From Förberedande kurs i matematik 1

If we examine the equation, we see that \displaystyle x only occurs as \displaystyle \sin x and it can therefore be appropriate to take an intermediary step and solve for \displaystyle \sin x, instead of trying to solve for \displaystyle x directly.

If we write \displaystyle t = \sin x and treat \displaystyle t as a new unknown variable, the equation becomes

\displaystyle 2t^2 + t = 1

and it is expressed completely in terms of \displaystyle t. This is a normal quadratic equation; after dividing by 2, we complete the square on the left-hand side,

\displaystyle \begin{align} t^2 + \frac{1}{2}t - \frac{1}{2} &= \Bigl(t+\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 - \frac{1}{2}\\[5pt] &= \Bigl(t+\frac{1}{4}\Bigr)^2 - \frac{9}{16} \end{align}

and then obtain the equation

\displaystyle \Bigl(t+\frac{1}{4}\Bigr)^2 = \frac{9}{16}

which has the solutions \displaystyle t=-\tfrac{1}{4}\pm \sqrt{\tfrac{9}{16}}=-\tfrac{1}{4}\pm \tfrac{3}{4}, i.e.

\displaystyle t = -\frac{1}{4}+\frac{3}{4} = \frac{1}{2}\qquad\text{and}\qquad t = -\frac{1}{4}-\frac{3}{4} = -1\,\textrm{.}

Because \displaystyle t=\sin x, this means that the values of x that satisfy the equation in the exercise will necessarily satisfy one of the basic equations \displaystyle \sin x = \tfrac{1}{2} or \displaystyle \sin x = -1\,.

\displaystyle \sin x = \frac{1}{2}:

This equation has the solutions \displaystyle x = \pi/6 and \displaystyle x = \pi - \pi/6 = 5\pi/6 in the unit circle and the general solution is

\displaystyle x = \frac{\pi}{6}+2n\pi\qquad\text{and}\qquad x = \frac{5\pi}{6}+2n\pi\,,

where n is an arbitrary integer.

\displaystyle \sin x = -1:

The equation has only one solution \displaystyle x = 3\pi/2 in the unit circle, and the general solution is therefore

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

where n is an arbitrary integer.

All of the solution to the equation are given by

\displaystyle \left\{\begin{align} x &= \pi/6+2n\pi\,,\\[5pt] x &= 5\pi/6+2n\pi\,,\\[5pt] x &= 3\pi/2+2n\pi\,, \end{align}\right.

where n is an arbitrary integer.