Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree. 4tan^2 θ+tanθ=−4tanθ−1

Question

Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree. 4tan^2 θ+tanθ=−4tanθ−1

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Calantha 3 years 2021-08-28T06:28:01+00:00 1 Answers 17 views 0

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  1. Jezebel
    0
    2021-08-28T06:29:02+00:00
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    Answer:

    135 and 315degrees

    Step-by-step explanation:

    Given the expression

    4tan²θ+tanθ=−4tanθ−1

    Equate to zero

    4tan²θ+tanθ+4tanθ+1 = 0

    4tan²θ+5tanθ+1 = 0

    Let x = tanθ

    4x²+5x+1 = 0

    Factorize;

    4x²+4x+x+1 = 0

    4x(x+1)+1(x+1) = 0

    4x+1 = 0 and x+1 = 0

    4x = -1 and x = -1

    x = -1/4 and -1

    Since x = tanθ

    -1 =tanθ

    θ = arctan(-1)

    θ = -45degrees

    Since tan is negative in the 2nd and fourth quadrant;

    In the second quadrant

    θ = 180 – 45 = 135degrees

    In the fourth quadrant;

    θ = 360 – 45 = 315degrees

    Hence the required angles are 135 and 315degrees

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