f(x) = x - c - dcos(x)

Case 1 dcos(c) > 0

f(c) = -dcos(c) < 0

f(c + pi) = pi -dcos(c+pi) = pi + dcos(c) > 0

We're done

Case 2 dcos(c) < 0

f(c) = -dcos(c) > 0

Let n be such that npi > dcos(c)

f(c-npi) = -npi -dcos(c-npi) = -npi - dcos(c)cos(npi)

cos(npi) =+1

in either case f(c-npi) < 0