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anonymous

  • one year ago

Solving trig equation on the interval 0<theta<2pi

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  1. anonymous
    • one year ago
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    \[3-3\sin \theta =6\cos^2\theta\]

  2. anonymous
    • one year ago
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    I solved it to equal \[0=6\sin^2\theta+3\sin \theta-2\] just need help factoring

  3. Astrophysics
    • one year ago
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    Let sin theta = x, and solve as you would regularly than plug it back in after :)

  4. anonymous
    • one year ago
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    That'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 :(

  5. Mertsj
    • one year ago
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    \[3-3\sin \theta =6(1-\sin ^2\theta)\] \[3-3\sin \theta =6-6\sin ^2\theta \] \[6\sin ^2\theta -3\sin \theta -3=0\]\[2\sin ^2\theta-\sin \theta-1=0\]

  6. Mertsj
    • one year ago
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    \[(2\sin \theta+1)(\sin \theta-1)=0\]

  7. Mertsj
    • one year ago
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    Set each factor equal to 0 and solve.

  8. anonymous
    • one year ago
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    ah I did \[3-3\sin \theta = 1+6\sin^2\theta\] 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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