From Förberedande kurs i matematik 1

If we use the formula for double angles, \displaystyle \sin 2x = 2\sin x\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\,\textrm{.}

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

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

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

\displaystyle \cos x = 0:

This equation has the general solution

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

\displaystyle 2\sin x-\sqrt{2}=0:

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

\displaystyle \left\{\begin{align} x &= \frac{\pi}{4}+2n\pi\,,\\[5pt] x &= \frac{3\pi}{4}+2n\pi\,, \end{align}\right.

where n is an arbitrary integer.

The complete solution of the equation is

\displaystyle \left\{\begin{align} x &= \frac{\pi}{4}+2n\pi\,,\\[5pt] x &= \frac{\pi}{2}+n\pi\,,\\[5pt] x &= \frac{3\pi}{4}+2n\pi\,, \end{align}\right.

where n is an arbitrary integer.