Answer
$\left\{\dfrac{3\pi}{2}\right\}$
Work Step by Step
We have to solve the equation:
$2\sin^2x-\sin x-3=0$, $0\leq x<2\pi$
Factor (or note $y=\sin x$ and use the quadratic formula):
$2\sin^2 x+2\sin x-3\sin x-3=0$
$2\sin x(\sin x+1)-3(\sin x+1)=0$
$(\sin x+1)(2\sin x-3)=0$
Solve for $x$:
$\sin x+1=0$ or $2\sin x-3=0$
$\sin x=-1$ or $\sin x=\dfrac{3}{2}$
$x=\dfrac{3\pi}{2}$ (the equation $\sin x=\dfrac{3}{2}$ has no solution as $\sin x\leq 1$
The solution set is:
$\left\{\dfrac{3\pi}{2}\right\}$
