Solution
= Solution
With $z=\cos x$,
$$
\frac{dz}{dx}=-\sin x,
\qquad
\frac{d^2z}{dx^2}=-\cos x=-z.
$$
The chain rule gives
$$
\frac{d^2w}{dx^2}
=\left(\frac{dz}{dx}\right)^2\frac{d^2w}{dz^2}
+\frac{d^2z}{dx^2}\frac{dw}{dz}
=(1-z^2)w_{zz}-zw_z.
$$
Hence $w_{xx}+n^2w=0$ becomes
$$
\boxed{(1-z^2)w_{zz}-zw_z+n^2w=0.}
$$