If cos(A) = cos(B), then

case 1: A = n*2Pi+B, where n is an integer,

case2: A = n*2Pi-B

A = 2x+pi/2, B = x-pi/2

case 1:

2x+pi/2 = n*2pi+x-pi/2

x + pi = n*2pi

x = pi(2n-1), or x = -pi for n=0

case 2:

2x+pi/2 = n*2pi-(x-pi/2)

2x+pi/2 = n*2pi-x+pi/2

3x = n*2pi

x = 2nPi/3, or x = 0 for n=0

There are, at least, 2 methods of solving the equation given.

First, let's solve the equation as a general one:

2x+ pi/2 = +/- arccos(cos(x- pi/2)) + 2*k*pi

Let's find the positive solution:

2x+ pi/2 = x- pi/2 + 2*k*pi

We'll move the unknown to the left side:

2x-x=-pi/2 - pi/2 + 2*k*pi

x=-2*pi/2 + 2*k*pi

x=-pi + 2*k*pi

**x= pi(2*k-1)**

Let's find now the negative solution:

2x+ pi/2 = -x + pi/2 + 2*k*pi

After moving the unknown to the left and simplifying the similar terms, we'll have:

2x+x = 2*k*pi

3x = 2*k*pi

**x = 2*k*pi/3**

**The solutions of the equation are: **

**x={pi(2*k-1);2*k*pi/3}**

**Another manner of solving would be to transform the difference of functions into a product, using the formula:**

**cos a-cos b= 2*sin(a+b)/2*sin(b-a)/2**