Trigonometry! #148

Let $u = sin(x)$ so that $2u^2 + 5u - 3 = (2u-1)(u+3) = 0 \Rightarrow u = \frac{1}{2}, -3$ . Since $-1 \le sin(x) \le 1,$ only $sin(x) = \frac{1}{2}$ is permissible. For the interval $x \in [0, 3\pi]$ , there are only 4 possible solutions: $\boxed{x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}}.$