$\sin x(\sin x+\cos x)\in \left[ \dfrac{1-\sqrt{2}}{2}, \dfrac{1+\sqrt{2}}{2} \right]$

$2$ integers lie in this interval, namely $a=0,1$.

If $a=0$,

$\sin x(\sin x+\cos x)=0$

$\Rightarrow \sin x=0$

or

$\sin x+\cos x=0$

$\Rightarrow \tan x=-1$

$x=-\pi ,0,\pi$

$3$ solutions

$x=-\frac{\pi }{4},\frac{3\pi }{4}$

$2$ solutions

If $a=1$,

$\sin x(\sin x+\cos x)=1$

${\sin }^{2}x+\sin x\cos x=1$

$2{\sin }^{2}x+2\sin x\cos x=2$

$1-\cos 2x+\sin 2x=2$

$\sin 2x-\cos 2x=1$

$\Rightarrow$ Squaring both sides,

$\sin 4x=0$

$4x=-4\pi ,-3\pi ,\dots ,3\pi ,4\pi$

$x= -\pi,\; -\dfrac{3\pi}{4},\; -\dfrac{\pi}{2},\; -\dfrac{\pi}{4},\; 0,\; \dfrac{\pi}{4},\; \dfrac{\pi}{2},\; \dfrac{3\pi}{4},\; \pi$

Checking in the original equation gives:

$x= -\dfrac{3\pi}{4},\; -\dfrac{\pi}{2},\; \dfrac{\pi}{4},\; \dfrac{\pi}{2}$

$4$ solutions

Total solutions $=3+2+4=9$

$\boxed{9\text{ solutions}}$