Hello!

As you probably know, there are many solutions of the equation `cos(w)=0.`

The general solution is `w=+-pi/2+2 k pi,`

where `k` is any integer. Without `+-` it may be written as two sequences,

`w_1=pi/2+2k pi` and `w_2=-pi/2+2k pi.`

In our problem `w=3z+pi,` so

`3z+pi=pi/2+2k pi` or `3z+pi=-pi/2+2k pi.`

These equations are linear for `z` and may be solved easily:

`z_1=-pi/6+(2k pi)/3` and `z_2=-pi/2+(2k pi)/3.`

This is the answer (remember that `k` is any integer).

`cos(3z+Pi) =0`

The steps in finding the general solution:

1. Firstly you need to find your reference angle, `alpha`

`alpha=cos^-1(0) = Pi/2`

2. The format for the general solution of the cos trig ratio:

The set of all solutions to `cos(x) = y`

is:

`x =± alpha +2kPi` or `x= -alpha +2kPi` , where k is an integer.

3. Applying the format to our solution

`(3z+Pi)=(pi/2)+2kpi`

`3z =(pi/2)-pi+2kpi`

`3z=(-pi/2)+2kpi`

`z= (-pi/6)+2k(pi/3)`

OR

``(3z+Pi)=(-pi/2)+2kpi`

`'3z =(-pi/2)-pi+2kpi'`

`3z=(-3*pi/2)+2kpi)`

` `

``z= (-pi/2)+2k(pi/3)`