find the number of solution of `sin^(2) x-sin x-1=0` in `[-2pi, 2pi]`.

Explanation

To solve the equation \( \sin^2 x - \sin x - 1 = 0 \) in the interval \([-2\pi, 2\pi]\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin^2 x - \sin x - 1 = 0 \]
This is a quadratic equation in terms of \( \sin x \).
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