`sin(2x)sin(x) = cos(x)` Find the exact solutions of the equation in the interval [0, 2pi).

`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`