Question #99b0f

1 Answer
May 18, 2017

theta=pi/6+2n,5pi/6+2nθ=π6+2n,5π6+2n where nn is a whole number.

Explanation:

First, set the two rrs equal to each other:

0=1-2sintheta0=12sinθ

Now solve for thetaθ:

-1=-2sintheta1=2sinθ

1/2=sintheta12=sinθ

theta=sin^-1(1/2)θ=sin1(12)

theta=pi/6+2n,5pi/6+2nθ=π6+2n,5π6+2n where nn is a whole number.