If `cosx=k` has exactly one solutions in `[0,2pi]` , then write the values of k .

Explanation

Given :
`cotx=k`
If `k=0`, then
Put `K=0`
`cosx=0`
`implies cosx=-cos(pi/2)`
`implies x=(2n+1)(pi/2),n in 2`
Now, `x=3(pi/2),5(pi/2),7(pi/2)----`
for `n=1`
If `k=1 ` then `cosx=1`
`implies cosx=cos0`
`implies x=2mpi , m in 2`
Now, `x=2pi,4pi,6pi,8pi,---`
`m=1,2,3,4-----`
If
` k=-1`
then `cosx=-1`
`implies cosx=cospi`
`x=2ppi pm pi , p in Z`
Now ,
`x=2ppi+pi`
`x=3pi,5pi,7pi-----`
When `p=1,2,3,4,------`
Clearly we can see that for `x=pi`
`cosx=k` has exactly one solution.
`k=-1`