will award best answer circular functions help?

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}

(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

:)>

There are calculators online, you know...