Consider r1(t)= <cos(t)+ 1/2, sin(t)> and r2(t)= <cos(t)- 1/2, sin(t)> . Do you see that r1 gives vectors that rotate around a circle of radius 1 with center at (-1/2, 0) and that r2 gives vectors that rotate around a circle of radius 1 with center at (1/2, 0)?
Those circles intersect at \left(0,\sqrt{3}/2\right) and \left(0, -\sqrt{3}/2\right) but if you set r1 and r2 equal, you get <cos(t)+ 1/2, sin(t)>= <cos(t)- 1/2, sin(t)> so that cos(t)+ 1/2= cos(t)- 1/2 which is not true for any t!
The problem is that the two circles reach the points of intersection at different t. If you think of these as two object moving around circular paths with t as time, then the two paths cross but the objects are at those points of intersection at different times.
You would do better to write <cos(t)+ 1/2, sin(t)>= <cos(s)- 1/2, sin(s)> so that you have two equations, cos(t)+ 1/2= cos(s)- 1/2 and sin(t)= sin(s) for the two unknown numbers. Obviously, s= t will satisfy the second equation but then the first is unsolvable. Since it is true that sin(\pi/2- x)= sin(x) s= \pi/2- t will also work. But then cos(s)= cos(\pi/2- t)= cos(t) and we have the same problem- the two cosines cancel leaving 1/2= -1/2 which is impossible.
It is also true that sin(x- \pi)= sin(x) so we could have s= \pi- t Then cos(t)+ 1/2= cos(s)- 1/2 becomes cos(t)+ 1/2= cos(\pi- t)- 1/2= -cos(t)- 1/2 or 2cos(t)= -1, cos(t)= -1/2 so that t= 2\pi/3 and s= \pi- 2\pi/3= \pi/3. Now we have <cos(t)+ 1/2, sin(t)>= <cos(2\pi/3)+ 1/2, sin(2\pi/3)>= <-1/2+ 1/2, -\sqrt{3}/2>= <0, -\sqrt{3}/2> and <cos(s)- 1/2, sin(s)>= <cos(\pi/3)- 1/2, sin(\pi/3)>= <1/2- 1/2, +\sqrt{3}/2>= <0, \sqrt{3}/2>.
