Answer
$$x \approx 3.237164881$$
Work Step by Step
$$\eqalign{
& 1 + {x^2}\sin x = 0;{\text{ }}\frac{\pi }{2} < x < \frac{{3\pi }}{2}{\text{ }} \cr
& {\text{From the graph we can see that the equation has infinitely}} \cr
& {\text{solutions}}{\text{, it is a periodic function}}{\text{.}} \cr
& {\text{For the interval }}\frac{\pi }{2} < x < \frac{{3\pi }}{2}{\text{ we have one solution}} \cr
& {\text{Let }}f\left( x \right) = 1 + {x^2}\sin x,{\text{ differentiating}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {1 + {x^2}\sin x} \right] \cr
& f'\left( x \right) = {x^2}\cos x + 2x\sin x \cr
& {\text{Using the Newton's Method }} \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}},{\text{ }}n = 1,2,3,...{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{1 + x_n^2\sin {x_n}}}{{x_n^2\cos {x_n} + 2{x_n}\sin {x_n}}}{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{From the graph we can see that the possible initial }} \cr
& {\text{approximation is }}{x_1} = 3.2,{\text{ substituting into }}\left( {\bf{2}} \right) \cr
& {x_{n + 1}} = \left( {3.2} \right) - \frac{{1 + {{\left( {3.2} \right)}^2}\sin \left( {3.2} \right)}}{{{{\left( {3.2} \right)}^2}\cos \left( {3.2} \right) + 2\left( {3.2} \right)\sin \left( {3.2} \right)}} \cr
& {x_2} \approx 3.237961846 \cr
& {\text{Continuing the process we obtain}} \cr
& {x_3} \approx 3.237165227 \cr
& {x_4} \approx 3.237164881 \cr
& {x_5} \approx 3.237164881 \cr
& {x_4} = {x_5}{\text{ into 9 decimal places}}{\text{, then the solution for the given}} \cr
& {\text{interval is }} \cr
& x \approx 3.237164881 \cr} $$
