Question 809615
Please help me with this Calculus II question: 
 Find all points of intersection of the given curves. (Assume 
 0 ≤ θ ≤ 2π.
Order your answers from smallest to largest θ. If an intersection occurs at the pole, enter POLE in the first answer blank.)
 r = 1 − cos θ, r = 1 + sin θ
<pre>
 1 &#8722; cos &#952; = 1 + sin &#952;  , since both = r
    &#8722;cos &#952; = sin &#952;
Divide both sides by cos &#952;
   {{{(-cos(theta))/cos(theta)}}} = {{{sin(theta)/cos(theta)}}}
   -1 = tan(&#952;)

   
So &#952; has two possible values where 0 &#8804; &#952; &#8804; 2&#960;. the tangent is negative
in Q2 and Q4 and the reference angle is 45° or {{{pi/4}}}, so &#952; is {{{3pi/4}}}
in Q2 or {{{7pi/4}}} in Q4.
 
So the smaller value of &#952; is {{{3pi/4}}}, and the value of r is gotten
by substituting {{{3pi/4}}} for &#952; in either of the original equations:
r = 1 &#8722; cos &#952; = 1 - cos{{{(3pi/4)}}} = 1 - {{{(-sqrt(2)/2)}}} = 1 + {{{sqrt(2)/2}}} = {{{(2+sqrt(2))/2}}}
So the first point is (r,&#952;) = ({{{(2+sqrt(2))/2}}}, {{{3pi/4}}})
The larger value of &#952; is {{{7pi/4}}}, and the value of r is gotten
by substituting {{{7pi/4}}} for &#952; in either of the original equations:
r = 1 &#8722; cos &#952; = 1 - cos{{{(7pi/4)}}} = 1 - {{{(sqrt(2)/2)}}} = 1 - {{{sqrt(2)/2}}} = {{{(2-sqrt(2))/2}}}
So the second point is (r,&#952;) = ({{{(2-sqrt(2))/2}}}, {{{7pi/4}}})

Edwin</pre>