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Math Help - Trignometric equation help.

  1. #1
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    Trignometric equation help.

    Solve each equation on the interval 0<= θ < 2n

    1) sinē θ - cosē θ = 1 + cos θ

    2) cos (2θ) + 6 sinē θ = 4


    Thanks in advance guys.
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  2. #2
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    Quote Originally Posted by cloudzer0 View Post
    Solve each equation on the interval 0<= θ < 2n

    1) sinē θ - cosē θ = 1 + cos θ

    2) cos (2θ) + 6 sinē θ = 4


    Thanks in advance guys.
    1) Replace \sin^2 \theta with 1 - \cos^2 \theta and re-arrange to get a simple quadratic in \cos \theta:

    \cos \theta (2 \cos \theta + 1) = 0.

    Now solve \cos \theta = 0 and 2 \cos \theta + 1 = 0.

    ------------------------------------------------------------------------------------------

    2) Replace \cos (2\theta) with 1 - 2 \sin^2 \theta and re-arrange to get a simple quadratic in \sin \theta:

    \sin^2 \theta = \frac{3}{4}.

    Now solve \sin \theta = \frac{\sqrt3}{2} and \sin \theta = -\frac{\sqrt{3}}{2}.
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  3. #3
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    Thank you for your help, I actually worked it through and figured out those problems myself. However, there are still some confusions going on which I've been looking at for a good while.

    Couple of other different problems.

    Problem:
    cos θ = sin θ

    My work:
    cos θ - sin θ = 0
    => (cosθ - sin θ)ē = 0
    => cosēθ - 2cosθsinθ + sinēθ =0
    => 1-2cosθsinθ = 0

    I'm pretty much stuck at this point. Did I do something wrong?

    *************
    Another problem which I seem to have trouble with is:
    (tan θ - 1)(sec θ - 1) = 0
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  4. #4
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    Quote Originally Posted by cloudzer0 View Post
    Thank you for your help, I actually worked it through and figured out those problems myself. However, there are still some confusions going on which I've been looking at for a good while.

    Couple of other different problems.

    Problem:
    cos θ = sin θ

    Mr F says: Divide both sides by \cos \theta: \, \tan \theta = 0 .....

    My work:
    cos θ - sin θ = 0
    => (cosθ - sin θ)ē = 0
    => cosēθ - 2cosθsinθ + sinēθ =0
    => 1-2cosθsinθ = 0

    I'm pretty much stuck at this point. Did I do something wrong?

    *************
    Another problem which I seem to have trouble with is:
    (tan θ - 1)(sec θ - 1) = 0

    Mr F says: Solve the two equations \tan \theta -1 = 0 \Rightarrow \tan \theta = 1 ...... and \sec \theta - 1 = 0 \Rightarrow \frac{1}{\cos \theta} - 1 = 0 \Rightarrow \frac{1}{\cos \theta} = 1 \Rightarrow \cos \theta = 1 .......
    ..
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  5. #5
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    1) cos (4θ) - cos (6θ) = 0

    ------------------

    Find real zeros. interval 0<= x < 2n

    2) f(x) = 4 cosē x - 1
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  6. #6
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    Quote Originally Posted by cloudzer0 View Post
    1) cos (4θ) - cos (6θ) = 0

    Mr F says: \Rightarrow \cos (4 \theta) = \cos (6 \theta).

    Now use the fact that the solutions to \cos \alpha = \cos \beta are \alpha = \beta + 2 n \pi \, and \alpha = - \beta + 2n \pi where n is an integer .......


    ------------------

    Find real zeros. interval 0<= x < 2n

    2) f(x) = 4 cosē x - 1

    Mr F says: Solve 0 = \cos^2 x - 1 over the given domain.
    ..
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