Question #99b0f

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Answer 1 · 2017-05-18T01:22:40

$θ = \frac{π}{6} + 2 n, 5 \frac{π}{6} + 2 n$ $theta=pi/6+2n,5pi/6+2n$ where $n$ $n$ is a whole number.

First, set the two $r$ $r$ s equal to each other:

$0 = 1 - 2 sin θ$ $0=1-2sintheta$

Now solve for $θ$ $theta$ :

$- 1 = - 2 sin θ$ $-1=-2sintheta$

$\frac{1}{2} = sin θ$ $1/2=sintheta$

$θ = {sin}^{- 1} (\frac{1}{2})$ $theta=sin^-1(1/2)$

$θ = \frac{π}{6} + 2 n, 5 \frac{π}{6} + 2 n$ $theta=pi/6+2n,5pi/6+2n$ where $n$ $n$ is a whole number.