From Förberedande kurs i matematik 1

If we use the formula for double angles, \displaystyle \text{sin 2}x=\text{2sin }x\text{ cos }x, and move all the terms over to the left-hand side, the equation becomes

\displaystyle 2\sin x\cos x-\sqrt{2}\cos x=0.

Then, we see that we can take a factor cos x out of both terms,

\displaystyle \cos x\left( 2\sin x-\sqrt{2} \right)=0

and hence divide up the equation into two cases. The equation is satisfied either if \displaystyle \text{cos }x=0\text{ } or if \displaystyle 2\sin x-\sqrt{2}=0.

\displaystyle \text{cos }x=0\text{ }: this equation has the general solution

\displaystyle x=\frac{\pi }{2}+n\pi ( \displaystyle n an arbitrary integer)

\displaystyle 2\sin x-\sqrt{2}=0: If we collect \displaystyle \text{sin }x on the left-hand side, we obtain the equation \displaystyle \text{sin }x\text{ }={1}/{\sqrt{2}}\;, which has the general solution

\displaystyle \left\{ \begin{array}{*{35}l} x=\frac{\pi }{4}+2n\pi \\ x=\frac{3\pi }{4}+2n\pi \\ \end{array} \right. ( \displaystyle n an arbitrary integer)

The complete solution of the equation is

\displaystyle \left\{ \begin{array}{*{35}l} x=\frac{\pi }{4}+2n\pi \\ x=\frac{\pi }{2}+n\pi \\ x=\frac{3\pi }{4}+2n\pi \\ \end{array} \right. ( \displaystyle n an arbitrary integer).