Jika f(x) = 1+2x, g(x) = cos(x+π/6) dan (fog)(x) = 0, maka nilai x di antara 0 dan 2π adalah

fungsi komposisi

fog(x)= 0
f { g(x)} = 0
f{ cos(x +π/6)} = 0
1 + 2 cos (x + π/6) = 0
2 cos (x + π/6) = -1
cos (x + π/6) = - 1/2
cos (x +π/6)  = cos 2/3 π  atau cos (x+π/6) =  = cos 4/3 π
x + π/6 = 2/3 π  atau x+ π/6 = 4/3 π
x = (2/3 - 1/2) π  atau x = (4/3 - 1/6)π
x = 1/6 π  atau x = 7/6 π