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Q.

The fundamental period of the function $f (x) = 4 \cos^{4} (\frac{x - π}{4 π^{2}}) - 2 \cos (\frac{x - π}{2 π^{2}})$ is equal to

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a

$2 π^{3}$

b

$4 π^{3}$

c

$π^{3}$

d

$3 π^{3}$

answer is D.

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Detailed Solution

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

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

Period of $f (x) = \frac{π}{\frac{1}{2 π^{2}}} = 2 π^{3}$

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The fundamental period of the function f(x)=4cos4x−π4π2−2cosx−π2π2 is equal to