`sin(4x) = -2sin(2x)` Find the exact solutions of the equation in the interval `[0, 2pi).`

By a double angle formula `sin(4x)=2sin(2x)cos(2x).`

Therefore our equation may be rewritten as

`2sin(2x)cos(2x)+2sin(2x)=0,` or

`sin(2x)(cos(2x)+1)=0.`

So `sin(2x)=0` or `cos(2x)=-1.`

The general solutions are `2x=kpi,` or `x=(kpi)/2,`

and `2x=-pi/2+2kpi,` so `x=-pi/4+kpi,` where `k` is any integer.

The roots in the interval `[0, 2pi)` are

`x=0,` `x=pi/2,` `x=pi,` `x=(3pi)/2,` `x=(3pi)/4,` `x=(7pi)/4.`

By a double angle formula `sin(4x)=2sin(2x)cos(2x).`

Therefore our equation may be rewritten as

`2sin(2x)cos(2x)+2sin(2x)=0,` or

`sin(2x)(cos(2x)+1)=0.`

So `sin(2x)=0` or `cos(2x)=-1.`

The general solutions are `2x=kpi,` or `x=(kpi)/2,`

and `2x=-pi/2+2kpi,` so `x=-pi/4+kpi,` where `k` is any integer.

The roots in the interval `[0, 2pi)` are

`x=0,` `x=pi/2,` `x=pi,` `x=(3pi)/2,` `x=(3pi)/4,` `x=(7pi)/4.`