Find the number of solutions of `sin^2x-sinx-1=0in[-2pi,2pi]`

Explanation

`sin^(2)x-sinx-1=0`
`rArr sin x=(1 pm sqrt(5))/2`
`=(1-sqrt(5))/2" "[sin x=(1+sqrt(5))/2 gt 1" not possible"]`
Hence, x can attain two values in `[0, 2pi]` and two more values in `[-2pi, 0]`. Thus, there are four solutions.