Aus Online Mathematik Brückenkurs 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 \text{sin }x\text{ } as a common factor which we can take out:

\displaystyle \text{sin }x\text{ }\left( \cos 3x-2 \right)=0

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

\displaystyle x=n\pi ( \displaystyle n 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 -\text{1 } and \displaystyle \text{1}, which gives that the largest value of \displaystyle \cos 3x-2 is \displaystyle -\text{1 }.

The solutions are therefore

\displaystyle x=n\pi ( \displaystyle n an arbitrary integer).