`sin(2x)-sin(x)=0 , 0<=x<=2pi`

using the identity `sin(2x)=2sin(x)cos(x)`

`sin(2x)-sin(x)=0`

`=2sin(x)cos(x)-sin(x)=0`

`=(sin(x))(2cos(x)-1)=0`

solving each part separately,

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

General solutions for sin(x)=0 are,

`x=0+2(pi)n, x=pi+2(pi)n`

solutions for the range `0<=x<=2pi` are,

`x=0 , x=pi , x=2pi`

Now solving 2cos(x)-1=0,

`2cos(x)=1`

`cos(x)=1/2`

General solutions for cos(x)=1/2 are,

`x=(pi)/3+2pin , x=(5pi)/3+2pin`

solutions for the range...

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`sin(2x)-sin(x)=0 , 0<=x<=2pi`

using the identity `sin(2x)=2sin(x)cos(x)`

`sin(2x)-sin(x)=0`

`=2sin(x)cos(x)-sin(x)=0`

`=(sin(x))(2cos(x)-1)=0`

solving each part separately,

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

General solutions for sin(x)=0 are,

`x=0+2(pi)n, x=pi+2(pi)n`

solutions for the range `0<=x<=2pi` are,

`x=0 , x=pi , x=2pi`

Now solving 2cos(x)-1=0,

`2cos(x)=1`

`cos(x)=1/2`

General solutions for cos(x)=1/2 are,

`x=(pi)/3+2pin , x=(5pi)/3+2pin`

solutions for the range `0<=x<=2pi` are,

`x=pi/3 , x=(5pi)/3`

Hence all the solutions are,

`x=0,pi,2pi,pi/3 , (5pi)/3`