Oxford University MAT 2019 Solved Papers

Use ${\cos }^2\mathit{x}+{\sin }^2\mathit{x}=1$. The equation becomes ${\sin }^3\mathit{x}+1-{\sin }^2\mathit{x}=0$. Write $\mathit{s}$ for $\sin \mathit{x}$. The cubic ${\mathit{s}}^3-{\mathit{s}}^2+1$ is $-1$ at $\mathit{s}=-1$, is 1 at $\mathit{s}=0$, and is 1 at $\mathit{s}=1$. The turning points are at $\mathit{s}=0$ and $\mathit{s}=2∕3$, and the value at $2∕3$ is $23∕27>0$. So this cubic has one (negative) root.
There are two solutions for $\sin \mathit{x}=\mathit{s}$ in the given range.