Trig
posted by Nada .
Determine the roots of the function f(x)=2cos(3x)1. Can you please show all the work so it is easier to understand, thank you!!

set 2cos(3x)  1 = 0
2cos(3x) = 1
cos 3x = 1/2
I will use degrees, we can always switch to radians later
3x = 60° or 3x = 300° by the CAST rule
x = 20° or x = 100°
the period of cos 3x = 360/3 or 120°
so by adding/subtracting 120° to any of our answers we can produce as many new answers as we want,
so for 0 ≤ x ≤ 360°
x = 20°, 140°,100° , 220° , 260°, 340°
or in radians
x = π/9, 7π/9, 5π/9, 11π/9 , 13π/9, 17π/9 
Ok so how did you get x to equal 20, 140,100........
And I also don't get where 3x=60 or 100 came from, can you please explain these parts by showing the math? Thank you 
I knew, or my calculator told me, that
cos 60° = 1/2 or cos 300° = 1/2
so 3x had to be 60° or 3x had to be 300°
well if 3x = 60 , what is x? (divide both sides by 3)
same for 3x = 300
x = 100
(on your calc.
enter
2nd
cos
.5
=
you should get 60 ) 
Srry I keep asking questions, but I noticed that you left out the negative sign on the 2 when you set the equation to zero, was that a mistake or were you suppose to do it like that?