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Math Help - Solve trig equation please help

  1. #1
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    Solve trig equation please help

    Solve:
    6sinsquared x + 5sinxcosx - 5cossquared x = 1

    just cant seem to get it, any help is GREATLY APPRECIATEDD

    thankuu !
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  2. #2
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    Let a=\sin x and b=\cos x, and move the 1 over.

    Then you have 6a^2+5ab-6b^2=0. That factors nicely. See if you can work from there.
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  3. #3
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    sorry i dont no if this is a really stupid question, feeling realli thick today
    but wat do u mean wen u say move the 1 over ?
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  4. #4
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    I think he actually made a mistake.

    If you let a=\sin x and b=\cos x

    You get 6a^2 +5ab-5b^2 = 1 or 6a^2 + 5ab - 5b^2 -1 = 0 which do not factorise nicely.
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  5. #5
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    yea..thats wat i thought..i donno
    please any one if you can help ?
    thanksss
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  6. #6
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    I used a program to simplify the original equation, and it came out with:
    1+11\cos 2x = 5 \sin 2x

    I tried working it out like this first:
    Let's fisrt get rid of the -5\cos^2 x:
    6\sin^2 x + 5\sin x\cos x = 1 + 5\cos^2 x
    6\sin^2 x + 5\sin x\cos x = 1 + 5(1-\sin^2 x)
    6\sin^2 x + 5\sin x\cos x = 1 + 5-5\sin^2 x
    11\sin^2 x + 5\sin x\cos x = 6
    \sin x (11\sin x + 5\cos x) = 6
    11\sin x + 5\cos x = 6\csc x

    But it didn't give anything.

    Then I found that the calculator finished up with 2x, so I gave a go to double angle formulas.

    6\sin^2 + 5\sin x\cos x - 5\cos^2 x = 1
    6\sin^2 + 5\sin x\cos x - 5\cos^2 x = \sin^2 x + \cos^2 x
    5\sin^2 + 5\sin x\cos x - 5\cos^2 x = \cos^2 x
    5(\sin^2  - \cos^2 x + \sin x\cos x) = \cos^2 x
    5[-(\cos^2 - \sin^2 x) + \frac{1}{2}(2\sin x\cos x)] = \cos^2 x
    5(-\cos 2x +  \frac{1}{2}\sin 2x) = \cos^2 x
    -5\cos 2x +  \frac{5}{2}\sin 2x = \cos^2 x
    -10\cos 2x +  5\sin 2x = 2\cos^2 x

    Except I think I might have done an error (graph of step 2 and step 3 are different, dunno why)
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  7. #7
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    well thanx for taking da time to help me , wow i dont realli get wat uve done, looks extremely complicated, i dont think the question is meant to be that complicated because i can do the qs before and afta it
    if it helps the solution to the question is 35 degrees 9 minutes, 120 degrees 24 minutes, 215 degrees 9 minutes and 300 degrees and 24 minutes
    i jst dont no how to get there
    thanx tho !
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  8. #8
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    Well, what I've done is using the trig identities (mainly: \sin^2 x + \cos^2 x = 1)

    I also used the double angle formulas:
    \sin(A \pm B) = \sin A\cos B \pm \sin B\cos A
    \cos(A \pm B) = \cos A\cos B \mp \sin A\sin B
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  9. #9
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    Hi iiarthero,

    the solution is attached.
    Attached Files Attached Files
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  10. #10
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    thanx for ur help !!!!!
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