For invertible of f, f must be bijective (i.e, one-one onto).
If `x_(1), x_(2) in R`,
then `f(x_(1))=f(x_(2))`
`implies cos(5x_(1)+2)=cos(5x_(2)+2)`
`implies 5x_(1)+2=2npipm(5x_(2)+2)`
`implies x_(1)nex_(2)`
`therefore` f is not one-one.
But `-1lecos(5x+2)le1`
`therefore -1lef(x)le1`
Range `=[-1,1]subR`
`therefore` f is into mapping.
Hence, the function f(x) is no bijective and so it is not invertible.
