What are the points of inflection, if any, of f(x)=xsin^2x  on [0,2pi]?

points corresponding to
$f ' ' \left(x\right) = 0 ,$
between
$x = 0.7 \mathmr{and} 0.8$
$x = 1. 1 \mathmr{and} 1.2$
$x = 2.3 \mathmr{and} 2.4$
$x = 2.5 \mathmr{and} 2.6$
$x = 3.0 \mathmr{and} 3.4$
$x = 4.0 \mathmr{and} 4.1$
$x = 5.4 \mathmr{and} 5.5$
$x = 5.5 \mathmr{and} 5.6$
correct to one decimal place

$x = \frac{\pi}{2}$
and
$x = 3 \frac{\pi}{2}$

Explanation:

Given: f(x)=xsin^2x
Evaluating the curvature,
$f ' \left(x\right) = x \left(2 \sin x\right) \cos x + {\sin}^{2} x$
=$2 x \sin x \cos x + {\sin}^{2} x$
=$x \sin 2 x + {\sin}^{2} x$
Since, $2 \sin x \cos x = \sin 2 x$
Thus, $f ' \left(x\right) = x \sin 2 x + {\sin}^{2} x$
$f ' ' \left(x\right) = x \cos 2 x \left(2\right) + \sin 2 x \left(1\right) + 2 \sin x \cos x$
=$2 x \cos 2 x + \sin 2 x + \sin 2 x$
Now, $f ' ' \left(x\right) = 2 x \cos 2 x + 2 \sin 2 x$
f''(x)=0
implies
$2 x \cos 2 x + 2 \sin 2 x = 0$
Taking $2 \cos 2 x$ common
$\cos 2 x \left(x + \tan 2 x\right) = 0$
Either $\cos 2 z = 0$
where, x is clearly $\frac{\pi}{2} \mathmr{and} 3 \frac{\pi}{2}$
or
$x + \tan 2 x = 0$
By trial and error,

Inspecting for values between 0 and 2pi in the intervals of 0.1pi
At $x = 0 , f ' ' \left(x\right) = 0$
$f ' ' \left(x\right) = 0 ,$
between
$x = 0.7 \mathmr{and} 0.8$
$x = 1. 1 \mathmr{and} 1.2$
$x = 2.3 \mathmr{and} 2.4$
$x = 2.5 \mathmr{and} 2.6$
$x = 3.0 \mathmr{and} 3.4$
$x = 4.0 \mathmr{and} 4.1$
$x = 5.4 \mathmr{and} 5.5$
$x = 5.5 \mathmr{and} 5.6$
correct to one decimal place
MY thanks for Fleur for reminding