Roots of Trigonometric Functions in an Interval

Ted123
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Homework Statement



This isn't really a question on its own, rather a step in the solution to another question:

How would I prove that y= A\cos x + B\sin x (A, B arbitrary constants) has at least n zeroes in the interval [\pi , \pi (n+1)] where n\in\mathbb{Z}\;?

(I don't need to be too explicit about it)

I was thinking state that A\cos x + B\sin x = \sqrt{A^2+B^2}\cos (x + \alpha) 0\leq\alpha \leq 2\pi and say something about periodicity and when sine and cosine are + or -...
 
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Should be it. Don't forget to mention that α = -arccos(A/sqrt(A²+B²)).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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