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Math Help - Trigonometry

  1. #1
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    Trigonometry

    Solve the following equations given that 0<θ<360

    a) (3cot^2)θ+1=5csc30
    b) tan^2θ=1/3
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by johntuan View Post
    Solve the following equations given that 0<θ<360

    a) (3cot^2)θ+1=5csc30
    3 \cot^2 \theta + 1 = 5 \csc 30

    \Rightarrow \cot  \theta = \pm \sqrt{\frac{5 \csc 30 - 1}3}

    \Rightarrow \cot  \theta = \pm \sqrt{\frac{\frac 52 - 1}3}

    now what?

    b) tan^2θ=1/3
    \tan^2 \theta = \frac 13

    \Rightarrow \tan \theta = \pm \frac 1{\sqrt{3}}

    now what?
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  3. #3
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    a) (3cot^2)θ+1=5csc30
    That's the one posted. Correct and re-write it.
    As posted, it defies any ammount of head-scratching.


    b) tan^2θ=1/3
    If that is
    tan^2(theta) = 1/3, then,
    Take the square root of both sides,
    tan(theta) = +,- 1 / sqrt(3)
    Plus or minus tan value. Meaning, in every quadrant, there should be theta.

    tan(theta) = 1 / sqrt(3) is for theta = 30deg. That's in the 1st quadrant.
    For the 2nd quadrant, theta = 180 -30 = 150 deg
    For the 3rd quadrant, theta = 180 +30 = 210 deg
    For the 4th quadrant, theta = 360 -30 = 330 deg
    So,
    theta = 30, 150, 210, 330 degrees.
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