Explain why the equation x- cosx =0 has at least one solution
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: Explain why the equation cos x = x has at least one solution?
There are several ways to look at this question.
First, at x=0, cos(x) is equal to 1, and x is equal to just 0 (since x=0=0). So at this point cos(x) > x. But the cosine function oscillates forever while the graph of y=x rises continuously to infinity. Therefore, at some point, the graph of y=x will have to "catch up" to the graph of y=cos(x).
You can look at it geometrically as well. If you draw a unit circle, when cos(x)=x, the arc length of the angle x will be equal to the x-coordinate.
If you want a more rigorous answer, if we define f(x) = cos(x) - x, this function takes on both positive and negative values. (f(0) = 1, f(2*Pi) = (1-2*Pi). Due to the intermediate value theorem, f(x) must somewhere take on the value 0, which means that cos(x) will equal x, since their difference is 0.
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There are several ways to look at this question.
First, at x=0, cos(x) is equal to 1, and x is equal to just 0 (since x=0=0). So at this point cos(x) > x. But the cosine function oscillates forever while the graph of y=x rises continuously to infinity. Therefore, at some point, the graph of y=x will have to "catch up" to the graph of y=cos(x).
You can look at it geometrically as well. If you draw a unit circle, when cos(x)=x, the arc length of the angle x will be equal to the x-coordinate.
If you want a more rigorous answer, if we define f(x) = cos(x) - x, this function takes on both positive and negative values. (f(0) = 1, f(2*Pi) = (1-2*Pi). Due to the intermediate value theorem, f(x) must somewhere take on the value 0, which means that cos(x) will equal x, since their difference is 0.
I HOPE IT HELP YOU
PLS PUT ME ON BRILLIANT LIST
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