Question

How do you solve `2cos(x/3)+sqrt2=0`?

Answer 1

`x = (9pi)/4 + 6kpi`
`x = (15pi)/4 + 6kpi`

Explanation:

`2cos (x/3) + sqrt2 = 0`
`cos (x/3) = -sqrt2/2`
Trig Table and unit circle -->
`x/3 = +- (3pi)/4`

a. `x/3 = (3pi)/4 + 2kpi`
`x = 9pi/4 + 6kpi`
b. `x/3 = (5pi)/4 + 2kpi`
(Arc `(-3pi)/4` is co-terminal to arc `(5pi)/4`)
`x = 15pi/4 + 6kpi`,