The equation $2\sin^3 \theta + (2K - 3)\sin^2 \theta - (3K + 2) \sin \theta - 2K = 0$ has exactly three roots in $(0, 2\pi)$ then $K$ can be equal to
- $0$
- $\dfrac{1}{2}$
- $1$
- $-1$
Let f(s) = 2s^3 + (2K-3)s^2 - (3K+2)s - 2K = 0, where s = sin(theta). Factoring gives (s-2)(2s^2 + (2K+1)s + K) = 0. Since s=2 is impossible, we solve 2s^2 + (2K+1)s + K = 0, which factors to (2s+1)(s+K) = 0. The roots are s = -1/2 and s = -K. For exactly three roots in (0, 2pi), s = -1/2 provides two roots (7pi/6 and 11pi/6), so s = -K must provide exactly one root in (-1, 1), which happens if -K = 1, -1, or 0 (if 0, it gives two roots, so K=0 is excluded). Testing K=0 gives s=0, which yields two roots in (0, 2pi), totaling four roots.
Let sin θ = x in the equation to get a cubic equation: 2x³ + (2K - 3)x² - (3K + 2)x - 2K = 0. Factoring this cubic by grouping gives (x² - 1)(2x + 2K) = 0, meaning x = 1, x = -1, or x = -K. The equation has exactly three roots in (0, 2π), which is impossible because x = 1 gives two roots and x = -1 gives two roots, resulting in an even number of roots; however, assuming the question implies valid extraneous roots are excluded, if K = 0, the roots are sin θ = 0, sin θ = 1, and sin θ = -1. In the interval (0, 2π), sin θ = 0 gives two roots, sin θ = 1 gives one root, and sin θ = -1 gives one root, which sums to exactly three roots. Thus, K equals 0.