Question

Question #3ded9

Answer 1

Let `f(x)=x-cos x`.

We know that `f` is continuous on `[0,pi/2]` since it is the difference of two continuous functions `x` and `cos x`, and

`f(0)=-1<0< pi/2=f(pi/2)`.

By Intermediate Value Theorem, there exists `c in (0,pi/2)` s.t.
`f(c)=c-cos(c)=0`, which means that `c=cos(c)`.

Hence, `x=cos x` has a solution `c in (0,pi/2)`.