Finding Points of Intersection for Polar Curves

jjeddy
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Homework Statement


I need to find the 2 points of intersection (in polar form) of the two curves.

I know just by looking that the origin will be one of the points, (0,0)

The Attempt at a Solution



I have approached this two different ways,

1. set them equal to each other and tried to simplify.Which approach should i use?
 
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OK, i used tanθ=sinθ/cosθ and i solved for θ

i can substitute back into get my corresponding (r,θ) r point and i should have my point right?
 
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Right. That gives you one of the points. I'm not quite sure how you came up with your answer of θ=π/4 , so it's difficult to suggest how you should come up with the second point.
 
Is that right?I think that is my polar coordinate where it intersects
 
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Your first post states that you need to find the 2 points of intersection.

For what other value of θ, is tan(θ) = 1 ? What is the period of the tangent function?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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