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r = 1 is a circle with radius 1 we know r always has to positive so tell me all the values of theta in (0, 2pi) such that r >= 0 1+ 2sin(theta) >= 0 sin(theta) >= -1/2 tell me the answer
What?\[1 + 2\sin(\theta) = 1\]
\[2\sin(\theta) = 0\]Find the solutions to theta.
\[\Theta = 0\]
Yeah, there are many solutions
how to find the intersection of the curve?
in this system of coordinates theta can lie only between 0 to 2pi so you can say theta = 0 is the required intersection
But \(2\pi\) is one too.
what are the intersections of the curve?
one of the point of intersection of the curve is (1,0), ?
If we can write the coordinates as \((\theta,r)\) then \((0,1)\) is the intersection.
Wait, do you have polar coordinates?!
@ParthKohli 2pi is just a depiction of one complete cycle so 2pi coincides with zero itself
how about the other intersections??
\((2n\pi,1)\) where \(n \) is the set of all integers
the other intersection is ( 2\Pi, 1)?
@ParthKohli i don`t think (2npi, 1) will do because polar coordinates have limits 1. r is positive 2. theta lies b/w 0 to 2pi (so we don`t talk of theta beyond them) point me out if i am wrong
Yeah, you are right. So are we talking about polar coordinates?
\[\theta \le 2\Pi ?\]
yes i think these are polar @yaho021 boundary conditions are always controversial so either you include 0 (which is done always) or 2pi(which i have never seen till now) 0 will be safe side because after all 0 or 2pi depict same condition