The equation $y=sinx\sin \left(x+2\right)-{\sin }^2(x+1)$ represents a straight line lying in:

$y=\sin (x+2)\sin x-{\sin }^2\left(x+1\right)$
$=\frac{1}{2}\left[2\sin \left(x+2\right)\sin x-2{\sin }^2\left(x+1\right)\right]$

$\left[\because 2\sin A\sin B\mathrm{ }=\cos \left(A-B\right)-\cos \left(A+B\right) \& \cos 2A=1-2{\sin }^{2}A\right]$
$=\frac{1}{2}\left[\cos \left(x+2-x\right)-\cos \left(2x+2\right)-\left(1-\cos 2\left(x+1\right)\right)\right]$
$=\frac{1}{2}\left[\cos 2-\cos \left(2x+2\right)-1+\cos \left(2x+2\right)\right]$
$=\frac{1}{2}\left[1-\cos 2\right]=-\frac{1}{2}2{\sin }^{2}1$
$=-{\sin }^{2}1($ constant and negative $)$
$$
Hence, the straight line lies in the third and fourth quadrants only.