I could spend some time today on the problem. Finally it wasn't that easy, but ain't hard neither. As I didn't have a calculator, it was too long, because without the calculator it's more complicated.
First, we notice that . From this, we know that our function is bounded by the 2 lines. Equaling them to 0, we find that for the first and for the second. This implies that has at least a zero between and , but as we only consider positive roots in the exercise, we can say that f has at least a zero on .
To be rigorous, I don't see anything else than finding the derivative of and to study its sign over . But before this, I will make a verification. As I assumed had a zero on , I can verify it multiplying by . If the result is negative, then we're right assuming it has at least a zero on . So which is negative.
When , and when , . Therefore we conclude that is increasing on and decreasing on , which implies that cross only once the x axis somewhere on .
Edit : I didn't understand wellyou could consider the coordinates of the turning points of the function f(x) = 4 sin(x) + 1 - x for x > a, where a > 0 and f(a) = 0 for the first time