Question

How do you use the quadratic formula to solve `3costheta+1=1/costheta` for `0<=theta<360`?

Answer 1

`64^@26; 140^@14; 219^@86; 295^@74`

Explanation:

`3cos t + 1 = 1/(cos t)`
Cross multiply, and bring the equation to standard form:
`3cos^2 t + cos t - 1 = 0`
Solve this quadratic equation for cos t by using the improved quadratic formula (Socratic Search).
`D = d^2 = b^2 - 4ac = 1 + 12 = 13`--> `d = +- sqrt13`
There are 2 real roots:
`cos t = -b/(2a) +- d/(2a) = - 1/6 +- sqrt13/6`
`cos t = (- 1 +- sqrt13)/6`

a. `cos t = (- 1 + sqrt13)/6 = 0.434`
Calculator and unit circle give -->
cos t = 0.434 --> arc `t = +- 64^@26`
The co-terminal of t = - 64.26 is `t = 360 - 64.26 = 295^@74`
b. `cos t = (-1 - sqrt13)/6 = - 0.767`
Calculator and unit circle --> arc `t = +- 140^@14`
The co-terminal of t = - 140.14 is `t = 360 - 140.14 = 219^@86`
Answers for (0, 360):
`64^@26; 140^@14; 219^@86; 295^@74`