Question

Which of the following angles is in the solution set of sec^2θ - 3secθ - 2 = 0 for 0° ≤ θ < 360°.
ans:0°,56°60°,74°

Answer 1

Are you sure its \[\sec^2θ - 3\secθ - 2 = 0\]or is it \[\sec^2θ - 3\secθ + 2 = 0\]

Answer 2

If its the second, you can easily solve it via factoring\[\sec^2θ - 3\secθ + 2 = 0\]\[(\sec \theta - 1)(\sec \theta - 2) = 0\]\[\sec \theta - 1 = 0 \rightarrow \sec \theta = 1\rightarrow \theta = \sec^{-1} 1 = 0 rads = 0 degrees\]\[\sec \theta - 2 = 0 \rightarrow \sec \theta = 2 \rightarrow \theta = \sec^{-1} 2 = \pi/3, 5\pi/3 radians= 60 \deg, 300 \deg\]

Answer 3

its sec^2θ - 3secθ - 2 = 0

Answer 4

If its the first, you have to treat the equation like a quadratic.
sec^2 θ − 3sec θ − 2=0
let x = sec theta
x^2 - 3x - 2 = 0\[x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a} = \frac {3 \pm \sqrt {17}}{2}\]\[x = \sec \theta = \frac {3 \pm \sqrt {17}}{2} \rightarrow \theta = \sec^{-1} \frac {3 \pm \sqrt {17}}{2}\]\[\theta = \sec^{-1} \frac {3 + \sqrt {17}}{2} \approx 1.286 rad = 73.69 degrees\]\[\theta = \sec^{-1} \frac {3 - \sqrt {17}}{2} = \pi i = 180 - 67.61 i degrees\]

Answer 5

Sorry the answers aren't any of the choices :( The question may have an error.

Answer 6

I don't know because I copied and pasted it right from the website.

Answer 7

According to wolfram, what I got should be correct.