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Math Help - Sin, Cos, Tan

  1. #1
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    Sin, Cos, Tan

    My quesion says:

    Solve this equation for 0<x<2(pi)

    4sinx cosx -2(square root)3sinx + 2cosx - (square root)3 = 0
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    Quote Originally Posted by onenameless View Post
    My quesion says:

    Solve this equation for 0<x<2(pi)

    4sinx cosx -2(square root)3sinx + 2cosx - (square root)3 = 0
    what you typed is confusing...

    i suppose you meant 4 \sin x \cos x - 2 \sqrt{3} \sin x + 2 \cos x - \sqrt{3} = 0

    in that case, factor out the common 2 \sin x from the first two terms:

    \Rightarrow 2 \sin x ( 2 \cos x - \sqrt{3}) + (2 \cos x - \sqrt{3}) = 0

    but now, 2 \cos x - \sqrt{3} is common to both terms, so pull that out

    \Rightarrow (2 \cos x - \sqrt{3})(2 \sin x + 1) = 0

    now continue...
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  3. #3
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    Yes, that was what i meant, sorry i could not find out how to type in a square root sign. But thanks, that's pretty much how much i needed.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by onenameless View Post
    Yes, that was what i meant, sorry i could not find out how to type in a square root sign. But thanks, that's pretty much how much i needed.
    so what's the solution?
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  5. #5
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    (2sinx + 1)(2cosx-squareroot3) = 0

    2sinx +1 = 0
    2sinx = -1
    sinx = -0.5
    x = -30

    2cosx - sr(3) = 0
    2cosx = sr(3)
    cosx = sr(3) / 2
    x = 30
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  6. #6
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    Quote Originally Posted by onenameless View Post
    (2sinx + 1)(2cosx-squareroot3) = 0

    2sinx +1 = 0
    2sinx = -1
    sinx = -0.5
    x = -30

    2cosx - sr(3) = 0
    2cosx = sr(3)
    cosx = sr(3) / 2
    x = 30
    your solutions are correct, but they are not the solutions we are after. the problem asked for 0 < x < 2 \pi, so your first solution is wrong, and your second is incomplete. there are two solutions for each (they coincide), can you find them?
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    Quote Originally Posted by Jhevon View Post
    your solutions are correct, but they are not the solutions we are after. the problem asked for 0 < x < 2 \pi, so your first solution is wrong, and your second is incomplete. there are two solutions for each (they coincide), can you find them?
    1) x = 210 and 330
    2) x = 30 and 150
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by onenameless View Post
    1) x = 210 and 330
    2) x = 30 and 150
    the one in red is wrong, the others are fine (remember what trig functions are positive in what quadrants)
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    the one in red is wrong, the others are fine (remember what trig functions are positive in what quadrants)
    But sin of x=30 is 0.5, and sin of x=150 is also 0.5, both are the same values with the same positive signs. 30 is in the first quadrant and 150 is in the second, which is where sin is always positive.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by onenameless View Post
    But sin of x=30 is 0.5, and sin of x=150 is also 0.5, both are the same values with the same positive signs. 30 is in the first quadrant and 150 is in the second, which is where sin is always positive.
    your second equation deals with cosine, not sine
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  11. #11
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    Quote Originally Posted by Jhevon View Post
    your second equation deals with cosine, not sine
    Oh man i did not see that. Right, then it's:
    x=30 and 330
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  12. #12
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    Quote Originally Posted by onenameless View Post
    Oh man i did not see that. Right, then it's:
    x=30 and 330
    correct
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