Find the peroid of the function f(x)=4cos^4((x-π)/(4π^2))-2cos((x-π)/…

$f(x)=4\cos^4(\frac{x-π}{4π^2})-2\cos(\frac{x-π}{2π^2})$

$=(2\cos^2(\frac{x-π}{4π^2}))^2-2\cos^2(\frac{x-π}{2π^2})$

$=(1+\cos(\frac{x-π}{2π^2}))^2-2\cos(\frac{x-π}{2π^2})$  $(using\,2\cos^2θ+\cos 2θ)$

$=\cos^2(\frac{x-π}{2π^2})+1$

$=\frac{1+\cos(\frac{x-π}{π^2})}{2}+1$

$=\frac{\cos(\frac{x}{π^2}-\frac{1}{π})}{2}+3$

for $\cos^2(\frac{x-π}{2π^2})+1$

we know for $\cos^2(x-π)+1$ period is $"π"$

so for $\cos^2(\frac{x-π}{2π^2})+1$

Period is $\frac{π}{\frac{1}{2}π^2}=2π^3$