Find the real roots of the equation `cos^4x + sin^7x = 1` in the interval `[-pi,pi]`

Explanation

We have,
`cos^(4)x + sin^(7)x=1`
`rArr sin^(7)x - (1-cos^(4)x)=0`
`rArr sin^(7)x-(1-cos^(2)x)(1+cos^(2)x)=0`
`rArr sin^(2)x[sin^(5)x-1-cos^(2)x]=0`
Either `sin^(2)x =0` or `sin^(5)x + sin^(2)x-2=0`
When `sin^(2)x=0 rArr x = npi, n int Z`
`rArr x=-pi, 0, pi` are real values of `x` in `[-pi,pi]` satisfying the given equations.
When `sin^(5)x + sin^(2)x-2=0`
`rArr (sinx-1)(sin^(4)x + sin^(3)x + sin^(2)x + 2sinx +2)=0`
`rArr sinx =1 rArr x=pi/2`, is the solution
Thus, real roots of the given equation are `x=pi/2, 0,pi,-pi`