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solve: sec^2(θ) - secθ - 2 = 0

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  • 6 months ago
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    Let's substitute u = sec θ

    So your equation is equal to:

    u² - u - 2 = 0

    Factor the left side:

    (u - 2)(u + 1) = 0

    Two possible solutions:

    u = 2

    or

    u = -1

    Now solve for:

    sec θ = 2

    or

    sec θ = -1

    There are infinitely many solutions to each of those. Were you given an interval for solutions?

    sec θ = 1/cos θ = 2

    cos θ = 1/2

    θ = arccos(1/2)

    θ = 2πn ± π/3, n ∈ ℤ

    sec θ = 1/cos θ = -1

    cos θ = 1/-1 = -1

    θ = arccos(-1)

    θ = 2πn ± π, n ∈ ℤ

    Answer:

    For the interval 0 ≤ θ < 2π

    θ = {π/3, π, 5π/3}

    For the interval -π < θ ≤ π

    θ = {-π/3, π/3, π}

    For the interval -π ≤ θ ≤ π)

    θ = {-π, -π/3, π/3, π}

  • 6 months ago

    (secθ + 1)(secθ - 2) = 0

    Then, either secθ = -1 or secθ = 2

    i.e. cosθ = -1 or cosθ = 1/2

    so, θ = ±3π/2 + 2nπ or ±π/3 + 2nπ...for n ∈ ℤ

    or, θ = (π/2)(4n ± 3) or (π/3)(6n ± 1)

    Hence, θ = π/3, π/2, 3π/2 and 5π/3...for 0 < θ < 2π

    :)>

  • 6 months ago

    There are calculators online, you know...

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