For what value of `k` the equation `sinx+cos(k+x)+cos(k-x)=2` has real solutions?

Explanation

`sinx+cos(k+x) + cos(k-x) = 2`

`=>sinx +2cos((k+x+k-x)/2)cos((k+x-k+x)/2) = 2`

`=>sinx+2coskcosx =2`

`=>sqrt(1+4cos^2k)(1/sqrt(1+4cos^2k)sinx+(2cosk)/sqrt(1+4cos^2k)cosx) =2`

Let `sin alpha = 1/sqrt(1+4cos^2k), then cos alpha = (2cosk)/sqrt(1+4cos^2k)`

So, our equation beecomes,

`=>sqrt(1+4cos^2k)(sinalphasinx +cosalphasinx) =2`

`=>sqrt(1+4cos^2k)cos(x-alpha) = 2`

...