Context: In the video, he divides out a variable as below: $$r^2=2rsin\theta\\r=2sin\theta$$ When you do a division like this, aren't you dividing out a variable? What happens if you take the square root of both sides like below? $$r^2=2rsin\theta\\|r|=\sqrt{2rsin\theta}$$ Would that be more correct?
Yes, he is. I particular, he has "lost" the r= 0 for all [itex]\theta[/itex] solution. I'm not sure what you mean by "more correct" for [itex]|r|= \sqrt{2r sin(\theta)}[/itex]. It is not at all a "solution" because you have not "solved for r".
Well, of course. The more important point is that you are potentially losing a solution of the equation; namely, r = 0. In this case, it doesn't matter because the equation r = 2sin(θ) still has r = 0 when θ = 0 + k##\pi##. No, and besides, what's the point? You haven't solved for r since it still appears on both sides of the equation.
Because that's the solution that always disappears when you divide by a variable. The only root of ##rA(...) = rB(...)## with ##A\neq{B}## is ##r=0##, and that's the one that we're losing when we divide both sides by ##r## to turn the equation into ##A(...) = B(...)##.
Any time you divide both sides of an expression by something, you should check to see whether that thing can be zero. If it can be, then you have lost solutions corresponding to when that equals zero. For example, if someone asks you to solve: (x-1)(x-2) = (x-1)(2x-1) The first thing we do is divide both sides by x-1 (x-2) = (2x-1) Now we solve for x x = -1 OK, but we divided both sides by x-1, and we have to be worried about when that is equal to zero. x-1= 0 when x=1, so I should go back to the original equation and check that x=1 is a solution, which it is easily seen to be.
Simply do: r^2=2rsin\theta\ <-> r(r-2sin\theta\)=0 <-> r = 0 or r = 2sin\theta\ problem solved ! (sorry I dont know how to do the maths letters here )