The period of the function f(x) = cos 4x tan 3x is given by the smallest positive value of T such that f(x + T) = f(x) for all x.

We can use the fact that the period of cos 4x is π/2 and the period of tan 3x is π/3. Then, we need to find the smallest positive value of T such that f(x + T) = f(x) for all x:

f(x + T) = cos 4(x + T) tan 3(x + T)
= cos(4x + 4T) tan(3x + 3T)
= cos 4x tan 3x + cos(4x + 4T) [tan(3x + 3T) - tan 3x]

Using the identity tan (a + b) = (tan a + tan b) / (1 - tan a tan b), we can simplify the second term as:

[tan(3x + 3T) - tan 3x] = [tan(3x + 3T) - tan(3x) + tan(3x) - tan(3x)] / [1 - tan(3x) tan(3x + 3T)]
= [3tanT + O(T^2)] / [1 - 3tan^2(3x) T - 3tan(3x) tanT - tan^2(3x) T]

where O(T^2) represents terms of order T^2 and higher that can be neglected for small values of T.

Substituting this back into the expression for f(x + T), we get:

f(x + T) = cos 4x tan 3x + cos(4x + 4T) [3tanT + O(T^2)] / [1 - 3tan^2(3x) T - 3tan(3x) tanT - tan^2(3x) T]

To simplify this expression, we can use the fact that cos(4x + 4T) = cos 4x cos 4T - sin 4x sin 4T and expand the denominator using the identity 1 - tan^2(3x) T = sec^2(3x) T - tan^2(3x) T.

After some algebraic manipulations, we get:

f(x + T) = cos 4x tan 3x + [3cos 4x sin 4T + cos 4T sin 4x] / [sec^2(3x) T - 3tan(3x) tanT - tan^2(3x) T]

To make f(x + T) equal to f(x), we need the numerator of the second term to vanish, which implies:

3cos 4x sin 4T + cos 4T sin 4x = 0

Using the identity sin (a + b) = sin a cos b + cos a sin b, we can rewrite this as:

2sin 4x (cos 4T - 3) + 2cos 4x sin 4T = 0

Dividing both sides by 2sin 4x and using the identity cos (a - b