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Math Help - Help?

  1. #1
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    Exclamation Help?

    Could someone please explain to me why in 6sinx = 5 - 8 cosx, x can only equal 9652' and 33652'? Why can't it also be 238' and 2638'???
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Could someone please explain to me why in 6sinx = 5 - 8 cosx, x can only equal 9652' and 33652'? Why can't it also be 238' and 2638'???
    Because when you plug those last two angles in the original equation, the equation is false.

    The first two make the original equation true.

    -------------
    6sinX = 5 -8cosX
    Square both sides,
    36sin^2(X) = 25 -80cosX +64cos^2(X)
    36(1 -cos^2(X)) = 25 -80cosX +64cos^2(X)
    100cos^2(X) -80cosX -11 = 0
    Using the quadradtic formula,
    cosX = 0.91962 or -0.11962

    When cosX = 0.91962 .......positive cosine value, so X is in the 1st or 4th quadrants,
    X = 23.1294 deg .....1st quadrant
    X = 336.8706 deg ....4th quadrant

    When cosX = -0.11962 ......negative cosine value, so X is in the 2nd or 3rd quadrants,
    X = 96.8702 deg .....2nd quadrant
    X = 263.1298 deg ....3rd quadrant

    Since X = 23.1294 deg and X = 263.1298 deg do not make the original equation true, then it means X is not in the 1st and 3rd quadrants.
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  3. #3
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    Is there any way of telling straight away which X would not hold the trig. equation true without subbing it into the trig. equation itself? (To save time during exams)
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  4. #4
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    Can anyone please explain?
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  5. #5
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    Quote Originally Posted by xwrathbringerx View Post
    Is there any way of telling straight away which X would not hold the trig. equation true without subbing it into the trig. equation itself? (To save time during exams)
    No, you have to check. Sometimes you may get extraneous solutions.
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