\displaystyle{\left(-{1},\frac{{{3}\pi}}{{2}}\right)} lies on the polar graph

The given equation of the polar curve is ${\displaystyle r=1+\cos \theta .}$
By substituing the point ${\displaystyle (-1,\frac{3\pi }{2})\in r=1+\cos \theta }$
${\displaystyle r=1+\cos \theta }$
${\displaystyle -1=1+\cos \left(\frac{3\pi }{2}\right)}$
${\displaystyle -1=1+0}$
${\displaystyle -1\ne 1}$
Therefore, the point ${\displaystyle (-1,\frac{3\pi }{2})}$ does not satisfy the equation of the polar curve ${\displaystyle r=1+\cos \theta .}$
Use online graphing calculator and draw the graph of ${\displaystyle r=1+\cos \theta }$ as shown below in Figure 1 and identify the point ${\displaystyle (-1,\frac{3\pi }{2}).}$

From Figure 1 it can be noted that the point ${\displaystyle (-1,\frac{3\pi }{2})}$ lies on the graph.
Consider the point ${\displaystyle (-1,\frac{3\pi }{2})}$ and change its radial coordinate to 1 and subtract ${\displaystyle \pi }$ from the directional coordinate to obtain the alternate representation of the point ${\displaystyle (-1,\frac{3\pi }{2}).}$
Therefore, the point ${\displaystyle (-1,\frac{3\pi }{2})}$ can also be represented by ${\displaystyle (1,\frac{3\pi }{2}-\pi )=(1,\frac{\pi }{2}).}$
By substituing the point ${\displaystyle (1,\frac{\pi }{2})\in r=1+\cos \theta ,}$
${\displaystyle r=1+\cos \theta }$
${\displaystyle 1=1+\cos \left(\frac{\pi }{2}\right)}$
${\displaystyle 1=1+0}$
${\displaystyle 1=1}$
Thus, the point ${\displaystyle (1,\frac{\pi }{2})}$ satisfies the polar equation ${\displaystyle r=1+\cos \theta .}$
And from Figure 1 it can be seen that the points ${\displaystyle (1,\frac{\pi }{2})\text{and}(-1,\frac{3\pi }{2})}$ are the same points.
Therefore, due to this multiple identity, the point ${\displaystyle (-1,\frac{3\pi }{2})}$ lies on the graph even if does not satisfy the polar equation ${\displaystyle r=1+\cos \theta }$