Show That the Equation 5x + cos x = 0 has Exactly One Real Root.?

Can anyone explain to me how to show this? Even a hint telling me what theorem or rule to use will help, thanks.

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Intermediate Value Theorem
Let f(x) = 5x + cosx
f(-π) = -5π – 1 < 0 f(0) = 1 > 0
f(-π) < 0 < f(0) Since f(x) is continuous on the real number line, f(x) is continuous, thus by Intermediate Value Theorem there exist at least one real root in interval (-π, 0) Now to prove that there is only one real root in the interval f'(x) = 5 - sinx f''(x) = -cosx Suppose that there is more than 1 root in (-π, 0) i.e 2. Then there is (2 + 1 = 3) turning points (-π, 0). f''(x) = 0 ==> -cosx = 0 ==> x = -π (not in the interval) By Rolle’s Theorem, there doesn’t exist more than 1 real root.

cos(x) = 1 – x^2/2 + x^4/24 – x^6/720 ….
So if 5x = – cos(x)
then 5x = -(1 – x^2/2 + x^4/24 – x^6/720 …)
If we get a value of x that is such that when you multiply it by 5 and it comes just under the plus x axis for example, then there can only be one real root.
This kind of equation (using my limited math) can be done using a Newtonian approximation. I think there are other ways but they are much beyond what I know.
We can get an approximation to start with by using the quadratic formula
– (1 – x^2/2) = 5x
– 1 – 5x + x^2/2 = 0
the two values you get for this are -0.196 and +10.1
The only possible answer is -0.196
or thereabouts.
10.1 is too big and it is plus.

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