Question

How do you solve `Cos(pi/2 * x - 1) = 0`?

Answer 1

`x =(2 + (1 + 4 k) pi)/pi, k = {0,pm 1,pm 2,..}`

Explanation:

The equation `cos(theta)=0` verifies for all `theta = pi/2+2k pi` with `k = {0,pm 1,pm 2,..}`
Then `pi/2x-1` must be of the form `pi/2+2k pi`. Equating
`pi/2x-1=pi/2+2k pi` Solving for `x` we get
`x =(2 + (1 + 4 k) pi)/pi, k = {0,pm 1,pm 2,..}`