How do I find the exact solutions algebraically for the following equation?
sin 2x sin x = cos x
Any help is appreciated.
Without knowing the domain you could potentially be dividing by 0 when cos(x)=0 if you cancel. It is better to take cos(x) from both sides and factor out cos(x)
$\displaystyle 2sin^2(x)cos(x)-cos(x)=0$
$\displaystyle
cos(x)(2sin^2(x)-1)=0$
Either $\displaystyle cos(x)=0$ or $\displaystyle 2sin^2(x)-1=0$
This should give 3 principal solutions and there may be more or less depending on the domain (usually $\displaystyle 0 \leq x \leq 2\pi $)