Consider f(x)=3x+1-sinx. This function is continuous and infinitely differentiable on `RR`.
`f(0)=1gt0` and `f(-pi)=1-3pilt0.`
By the Intermediate Value Theorem there is at least one `c in (-pi, 0)` for which `f(c)=0.` So our equation has at least one solution.
If it has one more solution, by Rolle's Theorem there is `c_1` between roots such that `f'(c_1)=0.` But `f'(x)=3-cosx,` which is always >0. This contradiction proves that there is only one root.