From Förberedande kurs i matematik 1

If we examine the equation, we see that \displaystyle x only occurs as \displaystyle \text{sin }x\text{ } and it can therefore be appropriate to take an intermediary step and solve for \displaystyle \text{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

when it is expressed completely in terms of \displaystyle t. This is a normal second-degree equation; after dividing by \displaystyle \text{2}, we complete the square on the left-hand side,

\displaystyle \begin{align} & 2t^{2}+t-\frac{1}{2}=\left( t+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}-\frac{1}{2} \\ & =\left( t+\frac{1}{4} \right)^{2}-\frac{9}{16} \\ \end{align}

and then obtain the equation

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

which has the solutions \displaystyle t=-\frac{1}{4}\pm \sqrt{\frac{9}{16}}=-\frac{1}{4}\pm \frac{3}{4}, i.e. \displaystyle t=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2} and \displaystyle t=-\frac{1}{4}-\frac{3}{4}=-1

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

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

where \displaystyle n\text{ } is an arbitrary integer.

\displaystyle \text{sin }x=-\text{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 \displaystyle n\text{ } is an arbitrary integer.

All of the solution to the equation are given by

\displaystyle \left\{ \begin{array}{*{35}l} x={\pi }/{6}\;+2n\pi \\ x={5\pi }/{6}\;+2n\pi \\ x={3\pi }/{2}\;+2n\pi \\ \end{array} \right. ( \displaystyle n\text{ } an arbitrary integer)