From Förberedande kurs i matematik 1

After moving the terms over to the left-hand side, so that

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

we see that we can take out a common factor \displaystyle \cos x,

\displaystyle \cos x (\sqrt{2}\sin x-1) = 0

and that the equation is only satisfied if at least one of the factors \displaystyle \cos x or \displaystyle \sqrt{2}\sin x - 1 is zero. Thus, there are two cases:

\displaystyle \cos x=0:

This basic equation has solutions \displaystyle x=\pi/2 and \displaystyle x=3\pi/2 in the unit circle, and from this we see that the general solution is

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

where n is an arbitrary integer. Because the angles \displaystyle \pi/2 and \displaystyle 3\pi/2 differ by \displaystyle \pi, the solutions can be summarized as

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

where n is an arbitrary integer.

\displaystyle \sqrt{2}\sin x - 1 = 0:

If we rearrange the equation, we obtain the basic equation as \displaystyle \sin x = 1/\!\sqrt{2}, which has the solutions \displaystyle x=\pi/4 and \displaystyle x=3\pi /4 in the unit circle and hence the general solution

\displaystyle x=\frac{\pi}{4}+2n\pi\qquad\text{and}\qquad x=\frac{3\pi }{4}+2n\pi\,,

where n is an arbitrary integer.

All in all, the original equation has the solutions

\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.