Find the sum of the solutions to
\[2 \sin^3 x - 3 \sin x = -\frac{3}{2} \sin 2x\]in the interval $0 \le x \le 2 \pi.$
Question:
Find the sum of the solutions to
\[2 \sin^3 x - 3 \sin x = -\frac{3}{2} \sin 2x\]in the interval $0 \le x \le 2 \pi.$
Answer:
Let's think step by step.

user
5 \pi
Find the sum of the solutions to
\[2 \sin^3 x - 3 \sin x = -\frac{3}{2} \sin 2x\]in the interval $0 \le x \le 2 \pi.$
train