$2\sin^3 x=\sin x-\cos^2 x+1$. 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.
[Math] Solving third degree equation involving trigonometric functions
trigonometry
Best Answer
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.