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

Verified step by step guidance

1. 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: \( -2 \sin^{2} x - 3 \sin x - 1 = 0 \).
2. 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 \( 2y^{2} + 3y + 1 = 0 \).
3. 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 \( c=1 \).
4. 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 \( -1 \leq \sin x \leq 1 \).
5. 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.

Key Concepts

Trigonometric Equations

Trigonometric equations involve functions like sine, cosine, and tangent. Solving these equations means finding all angle values that satisfy the equation within a given interval, often requiring algebraic manipulation and use of identities.

Quadratic Form in Trigonometric Functions

Many trigonometric equations can be rewritten as quadratic equations by substituting a trigonometric function (e.g., sin x) with a variable. This allows the use of algebraic methods like factoring or the quadratic formula to find solutions.

Interval and Exact Solutions in [0, 2π)

When solving trigonometric equations, solutions are often restricted to a specific interval, here [0, 2π). Exact solutions refer to precise angle values, often expressed in terms of π, that satisfy the equation within this domain.