Question

How do you find the exact value of `sec^2theta-10sectheta+20=0` in the interval `0<=theta<2pi`?

Answer 1

`t = +- 82^@06`
`t = +- 68^@90`

Explanation:

`sec^2 - 10 sec + 20 = 0`
Solve this quadratic equation for sec x by using the quadratic formula in graphic form (Google Search):
`D = d^2 = b^2 - 4ac = 100 - 80 = 20` --> `d = +- 2sqrt5`
There are 2 real roots:
`sec t = -b/(2a) +- d/(2a) = 10/2 +- 2sqrt5/2 = 5 +- sqrt5`
Two values of cos t
`cos t = 1/(5 + sqrt5)` and `cos t = 1/(5 - sqrt5)`
a. `cos t = 1/7.24 = 0.138`
Calculator and unit circle give:
`t = +- 82^@06`
b. `cos t = 1/(5 - sqrt5) = 1/2.76 = 0.36`
`t = +- 68^@90`