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anonymous

  • 5 years ago

How do you solve 9sin^2(x)+6sin(x)cos(x)+4cos^2(x)=0? Thanks

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  1. anonymous
    • 5 years ago
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    Is it no solution?

  2. anonymous
    • 5 years ago
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    I'm trying to put 6sin(x)cos(x) into a different form consisting only of either sin(x) or cos(x), but I can only get 3sin(2x). :/ This one's tricky, but I don't know if it's definitely one with no solution.

  3. anonymous
    • 5 years ago
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    You can put 9-9cos^2(x) instead of 9sin^2(x) or 4-4sin^2(x) instead of 4cos^2(x), but I don't know how to break down the middle term.

  4. anonymous
    • 5 years ago
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    well, the middle term equals [(6/2) sin 2x]

  5. anonymous
    • 5 years ago
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    since: sin x * cos x = 1/2 sin 2x

  6. anonymous
    • 5 years ago
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    It can also become 5sin^2(x) - 3sin(2x) +4 but this seems more convoluted than before.

  7. anonymous
    • 5 years ago
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    Idk. I probably will need to ask my teacher tmr.

  8. anonymous
    • 5 years ago
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    It does seem more convoluted, mainly because of that damned 2x in the second term...sorry, I don't know what to tell you. :P Be sure to get back with the solution!

  9. anonymous
    • 5 years ago
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    Sure.

  10. anonymous
    • 5 years ago
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    Uh how about this set sinx = x, cosx = y you get this: \[0=9x^2+6xy+4y^2\] factor it and plug in subsitutions

  11. anonymous
    • 5 years ago
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    Do you have any ideas on how to factor it?

  12. anonymous
    • 5 years ago
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    I don't think quadratic equation or any equation that I can think of right now will help.

  13. anonymous
    • 5 years ago
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    heh, Im still playing around with the coefficients

  14. anonymous
    • 5 years ago
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    Binomial expansion perhaps? Only slightly manipulated, maybe..

  15. anonymous
    • 5 years ago
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    Oh, 4y^2 can be c and 6y can be b.

  16. anonymous
    • 5 years ago
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    The discriminant is negative. No solution. Thanks for the tip.

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