To find x if cos(2x+pi/2)=cos(x-pi/2).

If cosa = cos b, then a = 2npi+b or 2npi-b.

Therefore if cos(2x+pi/2)=cos(x-pi/2), then

(i) 2x+pi/2 = 2npi+x-pi/2 or (ii) 2x+pi/2 = 2npi-x+p/2.

2x-x = 2npi-pi.

x = (2n-1)pi, n = 0, 1, 2,.....

(ii) 2x+pi/2 = 2npi -x+pi/2.

2x+x = 2npi.

3x= 2npi.

x = 2npi/3, n =0, 1, 2,

Since the interval of admissible values for x is not specified, we'll put the range of admissible values as R.

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

We'll re-write the equation:

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

We'll subtract pi/2;

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

We'll subtract x:

x = 2k*pi - pi

x1 = pi(2k-1)

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

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

We'll eliminate pi/2 both sides:

2x + x = 2k*pi

3x = 2k*pi

x2 = 2k*pi/3

**The solutions of the equation are: {2k*pi/3}U{pi(2k-1)}.**