If `|cosx|^(sin^2x-3/2sinx+1/2)=1` then `x=`

Explanation

To solve an equation of the form `f(x)^(g(x)) =1, f(x) ge 0`, we must break into two cases, viz, `f(x)=1, f(x) ne 1`.
Thus restriction on `g(x)` doesn't either into our consideration.
When `f(x) ne 1, f(x)^(g(x)) =1 rArr g(x)=0`
Returning to our problem
Case-I
`|cosx|=1`
`rArr cosx =1 rArr x =2npi`
and `cosx =-1 rArr x = 2npi + pi`
Case- II:
`|cos x| ne 1`
`rArr sin^(2)x -3/2 sin x +1/2=0`
`rArr 2 sin^(2)x - 3sinx +1=0`
`rArr (2 sinx-1)(sin x-1)=0`
First factor gives `sinx = 1/2 rArr x = npi +(-1)^(n) pi/6, n int Z`
Second factor gives `sinx =1`, but this makes the base, `|cos x|`, zero. So this is to be discarded.
Thus the complete solution is
`x=2npi, (2n+1)pi, npi + (-1)^(n) pi/6, n int Z`