5
\includegraphics[max width=\textwidth, alt={}, center]{d4bec1a9-2d24-4cf8-9991-9ab61ddbc865-08_430_990_260_539} Th id ag am sto cn \(\mathrm { y } = \frac { \sin 2 \mathrm { x } } { \mathrm { x } + 2 }\) fo \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Tb \(x\)-co \(\dot { \mathrm { d } } \mathbf { a }\) te 6 th max mm \(\dot { \mathrm { p } } n M\) is d t ed y \(\alpha\).

1. Fid \(\frac { \mathrm { dy } } { \mathrm { dx } }\) ad th t \(\alpha\) satisfies th eq tin \(\tan 2 x = 2 x + 4\) [4]
2. Stw alch atin \(\mathbf { b }\) t \(\alpha\) lies b tweerfd nd
3. Use th iterati fo mu a \(x _ { n + 1 } = \frac { 1 } { 2 } \tan ^ { - 1 } \left( 2 x _ { n } + 4 \right.\) to id b \& le \(6 \alpha\) co rect tod cimal p aces. Gie th resu to each teratio od cimal places.