Solving third degree equation involving trigonometric functions

Question

$2\sin^3 x=\sin x-\cos^2 x+1$. Solve for $x$.

I was able to turn it into a quadratic equation, and obtain the answers of $90$, $210$, and $330$ degrees. But the equation has six zeroes.

Answer 1

Your equation is equivalent to $2\sin^3 x=\sin x+\sin^2 x$.

Noting that $\sin x=0$ provides a solution, you can divide by $\sin x$ to obtain a quadratic.

That should give you the additional solutions, $\sin x=1$ and $\sin x=-0.5$.

In the interval $[0^\circ, 360^\circ)$, $\sin x=0$ and $\sin x=-0.5$ have two solutions, and $\sin x=1$ has one solution.

Answer 2

There are $3$ possible values of $\sin(x)$, but each may correspond to more than one value of $x$ in the interval $0$ to $360$ degrees.