Can someone explain how to prove their is only one real root?

Question

Answer 1

f(x) = 2x + cosx; f'(x) = 2 - sinx > 0 for all real x. So, this function increasing.

Now, f(0) = 1 >0 and f(-1) = - 2 + cos(- 1) < 0. So on interval (- 1, 0) this function has 0 and because f(x) increasing function it can't have more zeros