The IVT says simply if f(x) is continuous on a closed interval [a,b] then

f(x) takes on every value between f(a) and f(b) at least once.

f(x) =cos(x) - x

f(0) = 1 > 0

f(pi) = -1-pi < 0

Therfore f(x) =cos(x) - x has a zero 0n [0,pi]

i.e there is an x st cos(x)- x = 0

or cos(x) = x