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Math Help - Solve for θ

  1. #1
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    Solve for θ

    Solve for all + values of θ. If 0<θ<270

    3 - sec^2 2θ - tan^2 2θ = 0
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  2. #2
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    Quote Originally Posted by reiward View Post
    Solve for all + values of θ. If 0<θ<270

    3 - sec^2 2θ - tan^2 2θ = 0
    Remember that \sec^2{x} = \tan^2{x} + 1.


    So

    3 - (\tan^2{2\theta} + 1) - \tan^2{2\theta} = 0

    2 - 2\tan^2{2\theta} = 0

    2 = 2\tan^2{2\theta}

    1 = \tan^2{2\theta}

    \pm 1 = \tan{2\theta}

     2\theta = \left\{45^{\circ}, 135^{\circ}, 225^{\circ}, 315^{\circ}\right\} + 360^{\circ}n, n \in \mathbf{Z}

    \theta = \left\{22.5^{\circ}, 67.5^{\circ}, 112.5^{\circ}, 157.5^{\circ}, 202.5^{\circ}, 247.5^{\circ}\right\} in the domain 0^{\circ} \leq \theta \leq 270^{\circ}.
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  3. #3
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    oh sorry, my equation was wrong

    it should be

    3 - sec^2 2θ - tan 2θ = 0

    can you answer again?
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  4. #4
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    Quote Originally Posted by reiward View Post
    oh sorry, my equation was wrong

    it should be

    3 - sec^2 2θ - tan 2θ = 0

    can you answer again?

    Well , they look the same

    3 - (\tan^2{2\theta} + 1) - \tan{2\theta} = 0

    \tan^2 2\theta+\tan 2\theta-2=0

     <br />
(\tan 2\theta-1)(\tan 2\theta+2)=0<br />
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