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Thread: solve this trigonometric equation: [(2sinx+1)/(cosx)]*(cosx-sinx)+2=0

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    solve this trigonometric equation: [(2sinx+1)/(cosx)]*(cosx-sinx)+2=0

    Solve this trigonometric equation:

    [(2sinx+1)/(cosx)]*(cosx-sinx)+2=0
    Last edited by stapel; 02-01-2017 at 11:47 AM. Reason: Copying instructions from subject line back into post.

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    Quote Originally Posted by spartas View Post
    [(2sinx+1)/(cosx)]*(cosx-sinx)+2=0
    What did you mean by "solve this trigonometric equation"?

    Are you trying to calculate the value of "x", from the given equation?

    Please share your work.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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    to find the x, it has to do with tangent half-angle formulas!

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    Quote Originally Posted by spartas View Post
    to find the x, it has to do with tangent half-angle formulas!
    Okay - What is that formula? What does it looklike?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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    i have to replace the sinx with sinx=2t/1+t^2 cosx=1-t^2/1+t^2
    now it looks like
    \[\frac{2\frac{2t}{1+t^{2}}+1}{\frac{1-t^{2}}{1+t^{2}}}\times(\frac{1-t^{2}}{1+t^{2}}-\frac{2t}{1+t^{2}})+2=0\]

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    Cool

    Quote Originally Posted by spartas View Post
    i have to replace the sinx with sinx=2t/1+t^2 cosx=1-t^2/1+t^2
    In future, kindly please include all of the instructions when you post the exercise. Thank you!

    Quote Originally Posted by spartas View Post
    now it looks like
    \[\frac{2\frac{2t}{1+t^{2}}+1}{\frac{1-t^{2}}{1+t^{2}}}\times(\frac{1-t^{2}}{1+t^{2}}-\frac{2t}{1+t^{2}})+2=0\]
    Okay, so you're left with just algebra to do. Use what you learned about complex fractions (here), solving polynomials (here), and the Quadratic Formula (here) to simplify and solve.

    . . .the equation you posted:

    . . . . .[tex]\dfrac{\left(2\frac{2t}{1\, +\,t^{2}}\,+\,1\right)}{\left(\frac{1\,-\,t^{2}}{1\,+\,t^{2}}\right)}\, \times\, \left(\dfrac{1\,-\,t^{2}}{1\,+\,t^{2}}\,-\,\dfrac{2t}{1\,+\,t^{2}}\right)\,+\,2\,=\,0[/tex]

    . . .combining w/ common denominators:

    . . . . .[tex]\dfrac{\left(\frac{4t\, +\, (1\, +\, t^2)}{1\, +\, t^2}\right)}{\left(\frac{1\, -\, t^2}{1\, +\, t^2}\right)}\, \times\, \left(\dfrac{1\, -\, t^2\, -\, 2t}{1\, +\, t^2}\right)\, +\, 2\, =\, 0[/tex]

    . . .doing some simplification:

    . . . . .[tex]\left(\dfrac{4t\, +\, 1\, +\, t^2}{1\, -\, t^2}\right)\, \times\, \left(\dfrac{1\, -\, 2t\, -\, t^2}{1\, +\, t^2}\right)\, +\, 2\, =\, 0[/tex]

    Multiply through by the denominators to clear the fractions, and so forth. If you get stuck in finding the four solutions, please reply showing all of your steps so far, starting with the ones displayed above. Thank you!

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