Question

How do you find all solutions of the following equation in the interval [0,2pi] of `3cos2x+cosx+2=0`?

Answer 1

`70^@53; 289^@47, 60^@; 300^@`

Explanation:

3cos 2x + cos x + 2 = 0
Replace cos 2x by `(2cos^2 x - 1)` -->
`3(2cos^2 x - 1) + cos x + 2 = 0`
`6cos^2 x - 3 + cos x + 2 = 0`
Solve this quadratic equation for cos x
`6cos^2 x + cos x - 1 = 0`
`D = d^2 = b^2 - 4ac = 1 + 24 = 25` --> `d = +- 5`
There are 2 real roots:
`cos x = - b/(2a) +- d/(2a) = -1/12 +- 5/12`
`cos x = - 1/12 + 5/12 = 4/12 = 1/3`
`cos x = - 1/12 - 5/12 = - 6/12 = -1/2`
a. `cos x = 1/3 --> x = +- 70^@53`
Co-terminal arc of x = - 70.53 --> x = 289^@47
b. `cos x = -1/2 ---> x = +- 60^@`
Co-terminal of `x = - 60 ---> x = 300^@`
Answers for `(0, 2pi)`:
`70^@53; 289^@47; 60^@; 300^@`