From Förberedande kurs i matematik 1

If we move everything over to the left-hand side,

\displaystyle \sin x\cos 3x-2\sin x=0

we see that both terms have \displaystyle \sin x as a common factor which we can take out,

\displaystyle \sin x (\cos 3x-2) = 0\,\textrm{.}

In this factorized version of the equation, we see the equation has a solution only when one of the factors \displaystyle \sin x or \displaystyle \cos 3x-2 is zero. The factor \displaystyle \sin x is zero for all values of x that are given by

\displaystyle x=n\pi\qquad\text{(n is an arbitrary integer)}

(see exercise 3.5:2c). The other factor \displaystyle \cos 3x-2 can never be zero because the value of a cosine always lies between \displaystyle -1 and \displaystyle 1, which gives that the largest value of \displaystyle \cos 3x-2 is \displaystyle -1.

The solutions are therefore

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