Solve each equation for exact solutions over the interval [0, 2π).
―2 sin² x = 3 sin x + 1

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Rewrite the given equation to standard quadratic form in terms of \( \sin x \). Start with the equation: \( -2 \sin^{2} x = 3 \sin x + 1 \). Move all terms to one side to get: \( -2 \sin^{2} x - 3 \sin x - 1 = 0 \).

Multiply the entire equation by \(-1\) to simplify the coefficients: \( 2 \sin^{2} x + 3 \sin x + 1 = 0 \). Now, let \( y = \sin x \) to rewrite the equation as \( 2y^{2} + 3y + 1 = 0 \).

Solve the quadratic equation \( 2y^{2} + 3y + 1 = 0 \) using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a=2 \), \( b=3 \), and \( c=1 \).

Find the two possible values of \( y = \sin x \) from the quadratic formula. Then, determine which of these values lie within the valid range for sine, which is \( -1 \leq \sin x \leq 1 \).

For each valid \( \sin x \) value, find the corresponding \( x \) values in the interval \( [0, 2\pi) \) by using the inverse sine function and considering the sine function's symmetry in the unit circle.

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