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

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To solve the equation \( \sin^2 x - \sin x - 1 = 0 \) and find the number of solutions in the interval \( [-2\pi, 2\pi] \), we can follow these steps: ### Step 1: Substitute \( y = \sin x \) We start by letting \( y = \sin x \). The equation then becomes: \[ y^2 - y - 1 = 0 \] ...
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