Precalculus, Chapter 5, 5.5, Section 5.5, Problem 8

sin(2x)sin(x)=cos(x) ,0<=x<=2pi
sin(2x)sin(x)-cos(x)=0
=2sin(x)cos(x)sin(x)-cos(x)=0
=cos(x)(2sin^2(x)-1)=0
=cos(x)(sqrt(2)sin(x)-1)(sqrt(2)sin(x)+1)=0
solving each part separately,
cos(x)=0 , (sqrt(2)sin(x)-1)=0 , (sqrt(2)sin(x)+1)=0
General solutions for cos(x)=0 are,
x=pi/2+2pin , x=(3pi)/2+2pin
solutions for the range 0<=x<=2pi are,
x=pi/2 , x=(3pi/2)
(sqrt(2)sin(x)-1)=0
sin(x)=1/sqrt(2)
General solutions are,
x=pi/4+2pin , x=(3pi)/4+2pin
solutions for the range 0<=x<=2pi are,
x=pi/4 , x=(3pi)/4
sqrt(2)sin(x)+1=0
sin(x)=-1/sqrt(2)
General solutions are,
x=(5pi)/4+2pin, x=(7pi)/4+2pin
solutions for the range 0<=x<=2pi are,
x=(5pi)/4 , x=(7pi)/4
Combine all the solutions,
x=pi/2 ,x=(3pi)/2 , x=pi/4 , x=(3pi)/4 , x=(5pi/4) , x=(7pi)/4