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Math Help - solving this equation

  1. #1
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    solving this equation

    sin x - square root of 3 x cos x=2

    no idea how to do it o.o
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  2. #2
    Super Member Bacterius's Avatar
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    Quote Originally Posted by alessandromangione View Post
    sin x - square root of 3 x cos x=2

    no idea how to do it o.o
    This is ambigious. Please pick the correct one :

    1) \ \ \sin{(x)} - \sqrt{3x \cos{(x)}} = 2

    2) \ \ \sin{(x)} - \sqrt{3x} \cos{(x)} = 2

    3) \ \ \sin{(x)} - \sqrt{3} x \cos{(x)} = 2
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  3. #3
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    I read it as

    \ \ \sin{(x)} - \sqrt{3} \times \cos{(x)} = 2
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  4. #4
    Super Member Bacterius's Avatar
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    Lol, didn't see the middle x was slightly shorter than the other ones
    Actually it isn't. So it still was ambigious

    To answer the question : consider putting all trigonometric functions on one side, put the square root and the constant on the other, and see what you can work out.

    \sin{(x)} - \sqrt{3} \times \cos{(x)} = 2

    Use the trigonometric identity :

    \sin{(x)} - \sqrt{3} \times \sqrt{1 - \sin^2{(x)}} = 2

    Now arrange the equation :

    \sin{(x)} - 2 = \sqrt{3} \times (1 - \sin^2{(x)})

    Square both sides :

    (\sin{(x)} - 2)^2 = 3 \times (1 - \sin^2{(x)})

    Rearrange this :

    (\sin{(x)} - 2)^2 = 3 - 3 \sin^2{(x)}

    Expand on the left hand side :

    \sin^2{(x)} - 4 \sin{(x)} + 4 = 3 - 3 \sin^2{(x)}

    Rearrange :

    \sin^2{(x)} - 4 \sin{(x)} + 3 \sin^2{(x)} = 3 - 4

    Rearrange :

    4 \sin^2{(x)} - 4 \sin{(x)} = -1

    Keep going by factorizing :

    4 ( \sin^2{(x)} - \sin{(x)}) = -1

    Finish this up :

    \sin^2{(x)} - \sin{(x)} = \frac{-1}{4}

    And ... MAGIC ! Quadratic equation :

    \sin^2{(x)} - \sin{(x)} + \frac{1}{4} = 0

    Can you finish now ? It should be fairly easy.
    Last edited by Bacterius; December 30th 2009 at 04:46 PM.
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  5. #5
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    thank uou so much for the help!
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