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Solving a trig equation

  1. Apr 10, 2007 #1
    I have to solve:

    2sinx = cos(x-60), for 0<=x<=360

    so far i expanded the cos part

    2sinx = cosxcos60 + sinxsin60

    as cos60 = 1/2 and sin60 = (sqrt3)/2

    2sinx = (1/2)cosx + ((sqrt3)/2)sinx

    sinx = cosx + sqrt3 sinx

    (1-sqrt3)sinx = cosx

    but here's where I'm stuck, i have 2 trig functions in one equation and therefore cannot solve.
  2. jcsd
  3. Apr 10, 2007 #2


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    This line is incorrect. You have multiplied both sides by 2 to remove the factor of 1/2 on the right hand side, so the left hand side should be premultipled by 4.
  4. Apr 10, 2007 #3
    ahh yes, how silly of me,

    but even now i have:

    (4 - sqrt3)sinx = cosx

    i need it all in sins or cos'
  5. Apr 10, 2007 #4


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    Or, you could remember that tanx=sinx/cosx
  6. Apr 10, 2007 #5
    but if i divide through by cosx, i'll loose that solution of x because the cos' will cancel, its an equation not an expression.
  7. Apr 10, 2007 #6


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    No you won't. There's only going to be one solution to that equation in the range specified.
  8. Apr 10, 2007 #7
    the answer in the back of the text book gave 2 solutions.
    I was told to never divide through with an equation, only with an expression

    perhaps you only got one solution because you divided through to solve.
    Last edited: Apr 10, 2007
  9. Apr 10, 2007 #8


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    Sorry, Im being really stupid! You don't lose a solution by dividing by cos(x), but of course the function tan(x)= 1/(4-sqrt(3)), is periodic with period 180 degrees. So, the solutions to this in the given range will be the principal value for arctan(1/(4-sqrt(3))) [the one given by your calculator], and this value with 180 added on.
  10. Apr 10, 2007 #9
    ok so arctan 1/(4-sqrt3) gives me my priciple 23.8, and then +180 to give 203.8, my secondary.

    cheers, im a bit weary about dividing through but im not gonna argue with the outcome!
  11. Apr 10, 2007 #10
    It's fine because you know that an x where cos(x) is zero can't possibly be a solution (otherwise you have 4 - sqrt(3) = 0).
  12. Apr 10, 2007 #11
    If you really wanted to, when you had sin on one side and cos on the other side, you could square both sides (and possibly introduce extraneous roots), and use an identity for (sinx)^2 or (cosx)^2, changing it to a quadratic type. The method above is much easier.
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