Given,
$$
y=a \cos (\sin 2 x)+b \sin (\sin 2 x)
$$
On differentiating both sides w.r.t. $x$, we get
$$
\begin{aligned}
y^{\prime}=-a & \sin (\sin 2 x) \cos 2 x \cdot 2 \\
& +b \cos (\sin 2 x) \cdot(\cos 2 x) \cdot 2
\end{aligned}
$$
$$
y^{\prime}=2 \cos 2 x\{-a \sin (\sin 2 x)+b \cos (\sin 2 x)\}
$$
Again on differentiating, we get
$$
\begin{aligned}
y^{\prime \prime}=-4 \sin 2 x[-a \sin (\sin 2 x) & +b \cos (\sin 2 x)] \\
+ & 2 \cos 2 x[-a \cos (\sin 2 x) \cdot \cos 2 x \cdot 2 \\
& -b \sin (\sin 2 x) \cos 2 x \cdot 2]
\end{aligned}
$$
$$
\begin{aligned}
& =-4 \sin 2 x \cdot \frac{y^{\prime}}{2 \cos 2 x}-4\left(\cos ^2 2 x\right) y \\
& =-2(\tan 2 x) y^{\prime}-4\left(\cos ^2 2 x\right) y \\
\therefore y^{\prime \prime}+2(\tan 2 x) y^{\prime} & =-4\left(\cos ^2 2 x\right) y
\end{aligned}
$$