Question

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

Accepted Answer

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.

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

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Key Concepts

Trigonometric Equations

Quadratic Form in Trigonometric Functions

Interval and Exact Solutions in [0, 2π)