Question

How do you find 3cosx + cos2x = -2 ?

Answer 1

#x = (2k +1)pi;

`x = +- (2pi)/3 + 2kpi`

Explanation:

`3cos x + cos 2x + 2 = 0` (1)
Use trig identity
`cos 2x = 2cos^2 x - 1`
Replace in the equation cos 2x by `(2cos^2 x - 1)`
Equation (1) becomes a quadratic equation in cos x:
`2cos^2 x + 3cos x + 1 = 0.`
Since a - b + c = 0, use shortcut. The 2 real roots are:
cox = - 1 and `cos x = - c/a = - 1/2`
a. cos x = - 1
Unit circle gives --> `x = pi + 2kpi = pi(2k + 1)`
b. `cos x = - 1/2`
Trig table and unit circle give 2 solutions for x:
`x = +- (2pi)/3 + 2kpi`