We have to solve 3*cos(3x - 1) = 0 for x.

3*cos(3x - 1) = 0

=> cos(3x - 1) = 0

=> 3x - 1 = arc cos 0

=> 3x - 1 = pi/2 + 2*n*pi and 3x - 1 = 3*pi/2 + 2*n*pi

=> 3x = pi/2 +1 + 2*n*pi and 3x = 3*pi/2 + 1 + 2*n*pi

=> x = pi/6 +1/3 + 2*n*pi/3 and x = pi/2 + 1/3 + 2*n*pi/3

**The required solution is x = pi/6 +1/3 + 2*n*pi/3 and x = pi/2 + 1/3 + 2*n*pi/3**

First, we'll divide by 3 both sides:

cos(3x-1)=0

Now, we'll take the inverse function both sides. For this reason, since the cosine function is not one to one function and inverse function can only be applied to one to one functions, we'll restrict the domain of cosine function to [0;pi].

3x - 1 = +/-arccos (0) + 2kpi, k is an integer number.

3x - 1= +/- pi/2 + 2kpi

We'll add 1 both sides:

3x = +/- pi/2 + 2kpi + 1

We'll divide by 3 to isolate x:

x = +/- pi/6 + 2kpi/3 + 1/3

The solutions of the equation over the range [0;pi] are:

**x = pi/6 + 1/3 **