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anonymous
 one year ago
Solving trig equation on the interval 0<theta<2pi
anonymous
 one year ago
Solving trig equation on the interval 0<theta<2pi

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[33\sin \theta =6\cos^2\theta\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I solved it to equal \[0=6\sin^2\theta+3\sin \theta2\] just need help factoring

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Let sin theta = x, and solve as you would regularly than plug it back in after :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's what I'm having problems with, the factoring part. Whether it's sin theta or x doesn't make a difference to me. I have a lot of problems factoring :(

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.1\[33\sin \theta =6(1\sin ^2\theta)\] \[33\sin \theta =66\sin ^2\theta \] \[6\sin ^2\theta 3\sin \theta 3=0\]\[2\sin ^2\theta\sin \theta1=0\]

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.1\[(2\sin \theta+1)(\sin \theta1)=0\]

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.1Set each factor equal to 0 and solve.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah I did \[33\sin \theta = 1+6\sin^2\theta\] then got \[0=6\sin^2\theta+3\sin \theta 2\] I didnt realize you were supposed to write it out like that which made the rest of it get messed up. Thanks.
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