Question

Does `x^3+sin x = 1` have a root in `(0, 2pi)` ?

Answer 1

Yes, it has one root in `(0, 2pi)`

Explanation:

Let `f(x) = x^3+sin x - 1`, so zeros of `f(x)` are roots of our equation.

Note that:

• `f(0) = 0+0-1 = -1 < 0`

• `f(pi/2) = pi^3/8+1-1 = pi^3/8 > 0`

• Both `x^3` and `sin x` are continuous and strictly monotonically increasing in `(0, pi/2]`, so `f(x)` is too.

• `pi/2 ~~ 1.57 > root(3)(2)`, so when `x >= pi/2`, `f(x) > 2+sin(x)-1 > 0`

Hence by Bolzano's theorem `f(x)` has exactly one zero in `(0, 2pi)`, actually somewhere in `(0, pi/2)`.

graph{x^3+sinx -1 [-7, 7, -20, 20]}