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Trigonometric Equations Problem

  1. Aug 21, 2008 #1
    1. The problem statement, all variables and given/known data

    5sin(x) + (2 * squrroot 3) = (squarroot 3) + 3sin(x)

    2. Relevant equations
    sin (60) = square root 3 / 2

    C.A.S.T. rule

    3. The attempt at a solution

    5sin(x) + (2 * squrroot 3) = (squarroot 3) + 3sin(x)

    5sin(x) - 3sin(x) = (squarroot 3) - (2 * squrroot 3)

    2sin(x) = - (squarroot 3)

    sin(x) = - (squarroot 3) / 2

    Now....sin(x) = - (squarroot 3) / 2

    is 60 degrees by reference angle.

    However, because the sign was in a negative we must select the quadrants in which sine is negative. Those are the third and fourth quadrants. This means that the degrees will be 300 degrees and 240 degrees.

    However, the answer in my course material says the answers are 60 degrees and 120 degrees. But these are in quadrants in which the sine is positive...and as we saw in that equation, the sine is negative.

    Am I misreading something?

    Thanks all in advance.
  2. jcsd
  3. Aug 21, 2008 #2
    Hmm pretty simple equation. I don't see anything wrong with it. 240 deg and 300 deg seem like the correct answers for [tex]0 \leq x \leq 360[/tex] degrees.

    Maybe the 2sqrt(3) and sqrt(3) should be switched in the equation?
  4. Aug 23, 2008 #3
    You are correct here if the equation you stated is exactly what is stated in the book.

    Now if you did switch those so that equation would be:

    5sin(x) + \sqrt3 = 2\sqrt3 + 3sin(x)

    Then the answers in your book would be correct.

    Either you copied the wrong equation, or your book is incorrect (which is quite possible).
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