Question

How do you solve `5sin2x-6sinx=0` for `0<=x<=2pi`?

Answer 1

Solutions:
`x=0, x=pi,x=2pi,xapprox.927295,xapprox5.35589`

Explanation:

`5sin2x-6sinx=0`

First, use the [double angle identity](http://www.purplemath.com/modules/idents.htm`double) to simplify the`sin2x#: #5(2sinxcosx)-6sinx=0#

From here, we'll simplify and factor:
`10sinxcosx-6sinx=0`
`(2)(sinx)(5cosx-3)=0`

Now set each factor (that includes x) equal to zero:
`sinx=0`
`x=0`
`x=pi`
`x=2pi`

`5cosx-3=0`
`xapprox.927295`
`xapprox5.35589`