Solve 2 sin^2 t + sin t = 0 exactly on the interval from 0 to 2 pi.

1. Factor the common sine to get sin t (2 sin t + 1) = 0. 2. The first factor gives sin t = 0, which happens at t = 0 and t = pi within the interval. 3. The second factor gives 2 sin t = -1, so sin t = -1/2. 4. A negative sine puts the angle in quadrant III or IV, at 7 pi/6 and 11 pi/6. 5. Collecting all of them gives 0, pi, 7 pi/6 and 11 pi/6. 6. Four solutions appear even though a quadratic normally yields two, because each sine value can come from two different angles. 7. The angles pi/6 and 5 pi/6 would give sin t = +1/2, so that option has the sign of the second factor wrong. _Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 9 "Trigonometric Identities and Equations", section 9.5 Solving Trigonometric Equations_