18. How many solutions are there of the equation $1+2 \sin X-4 \sin ^{2} X-8 \sin ^{3} X=0$ with $0^{\circ}<X<360^{\circ}$ ?
A 1
B 2
C 4
D 6
E 8
Solution
C
For convenience we put $x=\sin X$.
We have
$$
\begin{aligned}
1+2 \sin X-4 \sin ^{2} X-8 \sin ^{3} X & =1+2 x-4 x^{2}-8 x^{3} \\
& =1+2 x-4 x^{2}(1+2 x) \\
& =(1+2 x)\left(1-4 x^{2}\right) \\
& =(1+2 x)(1+2 x)(1-2 x),
\end{aligned}
$$

It follows that
$$
\begin{aligned}
1+2 \sin X-4 \sin ^{2} X-8 \sin ^{3} X & =0 \\
& \Leftrightarrow(1+2 x)(1+2 x)(1-2 x)=0 \\
& \Leftrightarrow x=-\frac{1}{2} \text { or } x=\frac{1}{2} \\
& \Leftrightarrow \sin X=-\frac{1}{2} \text { or } \sin X=\frac{1}{2} .
\end{aligned}
$$

The solutions of the equation $\sin X=-\frac{1}{2}$ with $0^{\circ}<X<360^{\circ}$ are $x=210^{\circ}$ and $X=330^{\circ}$. The solutions of $\sin X=\frac{1}{2}$ in the same interval are $X=30^{\circ}$ and $X=150^{\circ}$.
Therefore the equation $1+2 \sin X-4 \sin ^{2} X-8 \sin ^{3} X=0$ has 4 solutions with $0^{\circ}<X<360^{\circ}$.