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Math Help - Trig Equations with Multiple Trig Functions cont.

  1. #1
    Member ~berserk's Avatar
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    Trig Equations with Multiple Trig Functions cont.

    Find, to the nearest tenth of a degree, all values of x in the interval 0≤x≤360 that satisfy the equation.

    tan x=cos x

    also solve for x in the interval 0≤x≤2π


    cos x= cos x
    Last edited by ~berserk; April 6th 2008 at 06:34 PM. Reason: edit
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by ~berserk View Post
    tan x=cos x
    \frac{sin x}{cos x} = cos x

    sin x = cos^2 x

    sin x = 1 - sin^2 x

     sin^2 x + sin x - 1 = 0

    Solve for x. (Let Sin x = k and solve using the quadratic formula.)

    Quote Originally Posted by ~berserk View Post
    cos x= cos x
     cos x - cos \frac{x}{2} = 0

    This is true where x is 0 and there also seems to be a solution between 45 and 60 degrees, as well as between 15 and 30 degrees. I suggest you draw an accurate graph to find the points.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    for the second one I suggest doing it this way

    cos\bigg(\frac{x}{2}\bigg)=cos(x)...now using the identity for cos\bigg(\frac{x}{2}\bigg) you obtain \sqrt{\frac{1+cos(x)}{2}}=cos(x)...now by squaring each side and simplifying you get 2cos(x)^2-1-cos(x)=0 Define u sin(x) and solve by quadratic formula
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  4. #4
    Member ~berserk's Avatar
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    i think i messed up the second problem yesterday because I did it and solved it like this

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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ~berserk View Post
    i think i messed up the second problem yesterday because I did it and solved it like this
    IDk...but here you go you factored it right so therefore cos(x)=1...so x=cos^{-1}(1)=0.... cos\bigg(\frac{1}{2{\cdot{0}/bigg)=cos(0)...and doing the other one you get x=\frac{2\pi}{3} which is a valid solution
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