Substituting \(y=\sum_{n=0}^{\infty} c_{n} x^{n}\) into the differential
equation we have $$\begin{aligned}
y^{\prime \prime}+(\sin x) y &=\sum_{n=2}^{\infty} n(n-1) c_{n}
x^{n-2}+\left(x-\frac{1}{6} x^{3}+\frac{1}{120}
x^{5}-\cdots\right)\left(c_{0}+c_{1} x+c_{2} x^{2}+\cdots\right) \\
&=\left[2 c_{2}+6 c_{3} x+12 c_{4} x^{2}+20 c_{5}
x^{3}+\cdots\right]+\left[c_{0} x+c_{1} x^{2}+\left(c_{2}-\frac{1}{6}
c_{0}\right) x^{3}+\cdots\right] \\
&=2 c_{2}+\left(6 c_{3}+c_{0}\right) x+\left(12 c_{4}+c_{1}\right)
x^{2}+\left(20 c_{5}+c_{2}-\frac{1}{6} c_{0}\right) x^{3}+\cdots=0
\end{aligned}$$ Thus $$\begin{array}{c}
2 c_{2}=0 \\
6 c_{3}+c_{0}=0 \\
12 c_{4}+c_{1}=0 \\
20 c_{5}+c_{2}-\frac{1}{6} c_{0}=0
\end{array}$$ and $$\begin{aligned}
c_{2} &=0 \\
c_{3} &=-\frac{1}{6} c_{0} \\
c_{4} &=-\frac{1}{12} c_{1} \\
c_{5} &=-\frac{1}{20} c_{2}+\frac{1}{120} c_{0}
\end{aligned}$$ Choosing \(c_{0}=1\) and \(c_{1}=0\) we find $$c_{2}=0, \quad
c_{3}=-\frac{1}{6}, \quad c_{4}=0, \quad c_{5}=\frac{1}{120}$$ and so on. For
\(c_{0}=0\) and \(c_{1}=1\) we obtain $$c_{2}=0, \quad c_{3}=0, \quad
c_{4}=-\frac{1}{12}, \quad c_{5}=0$$ and so on. Thus, two solutions are
$$y_{1}=1-\frac{1}{6} x^{3}+\frac{1}{120} x^{5}+\cdots \quad \text { and }
\quad y_{2}=x-\frac{1}{12} x^{4}+\cdots$$