Question

How do you find all solutions in the interval [0,2pi) of the equation `(cosx)^2 - 2.6 cosx - 0.87 =0`?

Answer 1

`107^@46 and 152^@54`

Explanation:

Solve the quadratic equation for cos x, using the improved quadratic formula.
`D = d^2 = b^2 - 4ac = 6.76 + 3.48 = 10.24`--> `d = +- 3.20`
`cos x = -b/(2a) +- d/(2a) = 1.3 +- 1.6`
cos x = 2.9 (rejected as > 1)
cos x = - 0.3 --> `x = +- 107^@46.`
Answer for `(0, 2pi)`:

`107^@46` and (360 - 107.46) = `252^@54` (co-terminal)