Trig Fns (1 Viewer)

[90atarpls]

hey guys, i cant get my head around this question, idk where to start..
solve for -pi (less than/equal to)x(less than/equal to)pi, 6cos^2(x) - 5cos(x) + 1 = 0

i really have no idea, and its qusetion 4 of the chapter review so must be fairly easy

 

[Audranda]

You have to factorise the equation:
6cos^2(x) - 5cos(x) + 1 = 0
(3cosx-1)(2cosx-1)=0
cosx= 1/3 cosx=1/2
x=... x=... (within the given domain)

 

[90atarpls]

You have to factorise the equation:
6cos^2(x) - 5cos(x) + 1 = 0
(3cosx-1)(2cosx-1)=0
cosx= 1/3 cosx=1/2
x=... x=... (within the given domain)

thx, i noticed that a few mins after i posted it, i shoulda just gave myself a little more time :S