= Solution
The chain rule gives $z_x=\sin2x$, $z_x^2=4z(1-z)$, and $z_{xx}=2(1-2z)$. Therefore
$$
4z(1-z)w_{zz}+2(1-2z)w_z+n^2w=0,
$$
which, after division, is the printed equation. It has only three <regular singular points>, $0,1,\infty$, and is thus a Papperitz equation. At zero and one the indicial roots are $0,1/2$. At infinity, inserting $w\sim z^{-\rho}$ gives $\rho=\pm n/2$. Its <Papperitz symbol> is
$$
\boxed{P\left\{\begin{matrix}0&\infty&1\\0&n/2&0\\1/2&-n/2&1/2\end{matrix};z\right\}.}
$$
The exponent sum is one, as required for a three-singularity second-order equation.
Multiplying the equation by $z(1-z)$ puts it in <Gauss hypergeometric equation> form with $a=n/2$, $b=-n/2$, $c=1/2$. The solution analytic at zero and normalized to one is ${}_2F_1(a,b;c;z)$. Composing it with $z=\sin^2x$ gives an even analytic solution of $w_{xx}+n^2w=0$ with $w(0)=1$, $w_x(0)=0$. Uniqueness of this <ordinary differential equation> implies
$$
\boxed{{}_2F_1(n/2,-n/2;1/2;\sin^2x)=\cos nx.}
$$
This <Gauss hypergeometric function> identity initially holds near zero, and on $|x|<\pi/2$ with the usual real branch; it extends by matching <analytic continuation>. It is not an identity on every real $x$ if the left side is always reevaluated on its principal $z$ branch: for $n=1$ that principal value is $\sqrt{1-\sin^2x}=|\cos x|$. Continuing the branch consistently restores the stated cosine.
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