Answer
$(1-\frac{\sqrt{2}}{2},\frac{3\pi}{4})$ and $(1+\frac{\sqrt{2}}{2},\frac{7\pi}{4})$
Work Step by Step
Eliminate $r$ or $\theta$:
$r=1+\cos\theta$ and $r=1-\sin\theta$
$r-1=\cos\theta$ and $r-1=-\sin\theta$
$(r-1)^2=\cos^2\theta$ and $(r-1)^2=\sin^2\theta$ (Substract both equations)
$0=\cos^2\theta-\sin^2\theta$
$0=\cos2\theta$
Find $\theta\in[0,2\pi]$ satisfying the last equation:
$\cos2\theta=0$
$\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},$ and $\frac{7\pi}{4}$
Back substitute $\theta$ above into the given polar equations:
$\theta=\frac{\pi}{4}$, $r_1=1+\frac{\sqrt{2}}{2}$ and $r_2=1-\frac{\sqrt{2}}{2}$ $\Rightarrow r_1\neq r_2$ (not satisfied)
$\theta=\frac{3\pi}{4}$, $r_1=1-\frac{\sqrt{2}}{2}$ and $r_2=1-\frac{\sqrt{2}}{2}$ $\Rightarrow r_1= r_2$ (satisfied)
$\theta=\frac{5\pi}{4}$, $r_1=1-\frac{\sqrt{2}}{2}$ and $r_2=1+\frac{\sqrt{2}}{2}$ $\Rightarrow r_1\neq r_2$ (not satisfied)
$\theta=\frac{7\pi}{4}$, $r_1=1+\frac{\sqrt{2}}{2}$ and $r_2=1+\frac{\sqrt{2}}{2}$ $\Rightarrow r_1= r_2$ (satisfied)
Thus, the intersection points are $(1-\frac{\sqrt{2}}{2},\frac{3\pi}{4})$ and $(1+\frac{\sqrt{2}}{2},\frac{7\pi}{4})$.