Solution 4.4:6a
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 \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).
