# Solve equation sin2x*sinx-1/2*sinx-sin2x=-1/2?

*print*Print*list*Cite

### 1 Answer

You may solve the given equation by factoring, hence, you need to bring all the terms to one side, such that:

`sin 2x*sin x - 1/2*sinx - sin2x + 1/2 = 0`

You need to form two groups, such that:

`(sin 2x*sin x - sin 2x) - (1/2*sinx - 1/2) = 0`

You need to factor out `sin 2x` in the first group, such that:

`sin 2x*(sin x - 1) - (1/2*sinx - 1/2) = 0`

You need to factor out `1/2` in the second group, such that:

`sin 2x*(sin x - 1) -(1/2)(sin x - 1) = 0`

Factoring out `sin x - 1` yields:

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

Using zero product rule, yields:

`sin x - 1 = 0 => sin x = 1 => x = (-1)^n*sin^(-1)(1) + npi`

`x = (-1)^n*pi/2 + npi`

`sin 2x - 1/2 = 0 => sin 2x = 1/2 => 2x = (-1)^n*sin^(-1)(1/2) + npi => 2x = (-1)^n*pi/6 + npi => x = (-1)^n*pi/12 + npi/2`

**Hence, evaluating the general solutions to the given equation, yields **`x = (-1)^n*pi/2 + npi, x = (-1)^n*pi/12 + npi/2.`

**Sources:**