on my exam review, i have this question for composition of functions

Given f(x)=3x^2+x-1, g(x)=2cos(x), determine the values of x when f(g(x))=1 for 0≤x≤2π.

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1. f(g(x))
= f(2cos(x) )
= 3(2cosx)^2 + 2cosx - 1
= 12 cos^2 x + 2cosx - 1

we want 12 cos^2 x + 2cosx - 1 = 1
12 cos^2 x + 2cosx - 2 = 0
6 cos^2 x + cosx - 1 = 0
(3cosx -1)(2cosx + 1) = 0
cosx = 1/3 or cosx = 1/2
since the domain is given in radians, our angle should be in radians.
Using my calculator, set to radians, I got

from cosx = 1/3
x = 1.23 or x = 2π-1.23 = 5.05
from x = 1/2
x = π/3 or x = 2π-π/3 = 5π/3

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posted by Reiny

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