The value x is equal to [tex]\pi[/tex] for which f(g(x))=g(f(x))
Given :
f(x)=3+cos(x)
g(x)=2x
To find : the value of x over the given interval [tex]0\leq x\leq 2\pi[/tex]
Apply composition of functions. Replace g(x) inside f(x) to find f(g(x))
[tex]f(g(x))=f(2x)=3+cos(2x)\\g(f(x))=g(3+cosx)=2(3+cosx)\\[/tex]
Now we set f(g(x))=g(f(x)) and solve for x
[tex]3+cos(2x)=2(3+cosx)\\3+cos(2x)=6+2cos(x)\\\cos \left(2x\right)-2\cos \left(x\right)-3=0\\\\we \; know:\ cos(2x)=2cos^2(x)-1\\-4-2\cos \left(x\right)+2\cos ^2\left(x\right)=0\\2cos^2(x)-2cos(x)-4=0\\\\divide \; by \; 2 \; and \; Factor \; left \; hand \; side\\\\cos^2(x)-cos(x)-2=0\\\\(cosx-2)(cosx+1)=0\\\cos \left(x\right)=2,\:\cos \left(x\right)=-1[/tex]
cos(x)=2 cannot be solved , so no solution
[tex]cos(x)=-1\\cos is -1 at x=180 degree\\So , x=\pi[/tex]
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