algebra 2 trig cumulative review 4 answers
2(\sin x)^2 - \sin x - 1 = 0 \). Let \( y = \sin x \), then solve \( 2y^2 - y - 1 = 0 \). Factor or use quadratic formula: \( y = \frac{1 \pm \sqrt{1^2 - 4 \times 2 \times (-1)}}{2 \times 2} \). Simplify discriminant: \( 1 + 8 = 9 \), so \( y = \frac{1 \pm 3}{4} \). Solutions for