Question

How do you solve `2cos^2x + 3cos x - 2 = 0` on the interval `[0,2pi]`?

Answer 1

Solve `f(x) = 2cos^2 x + 3cos x - 2 = 0.`

Ans: `+- pi/3`

Explanation:

Call cos x = t. Solve the quadratic equation:
`2t^2 + 3t - 2 = 0`
`D = d^2 = 9 + 16 = 25` --> `d = +- 5`
`t = cos x = -3/4 +- 5/4`
`t = 1/2` and t = -2 (rejected)
Trig table gives --> `cos x = 1/2` --> `x = +- pi/3`