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Chapter 5: Problem 40

Explain why the equation $$ (\cos x)^{99}+4 \cos x-6=0 $$ has no solutions.

Short Answer

Expert verified
The equation \((\cos x)^{99}+4 \cos x-6=0\) has no solutions because the polynomial \(f(\cos x) = (\cos x)^{99}+4 \cos x-6\) is continuous and differentiable within the valid range of the cosine function, \([-1,1]\). Evaluating the polynomial at -1 and 1, we get \(f(-1) = -11\) and \(f(1) = -1\), both of which are negative. Applying the Intermediate Value Theorem, we conclude that there are no values of \(\cos x\) within the valid range of the cosine function that will make the equation equal to zero.

Step by step solution

01

Identify the Domain of Cosine Function

The domain of the cosine function is all real numbers. However, it has a range, or a limited set of values it can take on. Specifically, the range of the cosine function is \([-1,1]\).
02

Identify the Properties of the Equation

The given equation is a polynomial in terms of cosine, which can be written as: \(f(\cos x) = (\cos x)^{99}+4 \cos x-6\) We must determine whether this polynomial has any zero within the range of the cosine function, that is, when \(\cos x \in [-1,1]\), find if there is any value that makes \(f(\cos x) = 0\).
03

Evaluate f(-1) and f(1)

We can start by evaluating the function at the endpoints of the possible cosine values, -1 and 1. For \(\cos x = -1\), the function becomes: \(f(-1) = (-1)^{99} + 4(-1) - 6 = -1 - 4 - 6 = -11\) For \(\cos x = 1\), the function becomes: \(f(1) = (1)^{99} + 4(1) - 6 = 1 + 4 - 6 = -1\)
04

Determine the Continuity and Differentiability

The function \(f(\cos x)\) is a polynomial, so it is continuous and differentiable everywhere in its domain (which in this case coincides with the whole range of the cosine function). This means that the function has no jumps, holes, or other discontinuities, and it has a continuous derivative.
05

Apply Intermediate Value Theorem

Since \(f(\cos x)\) is a continuous function on the interval \([-1, 1]\) and we have found that \(f(-1) < 0\) and \(f(1) < 0\), we can apply the Intermediate Value Theorem. The Intermediate Value Theorem states that if a continuous function is defined on a closed interval and takes two values of the same sign, then it cannot take the value zero within that interval. Since \(f(-1)\) and \(f(1)\) are both negative, according to the Intermediate Value Theorem, there are no values of \(\cos x\) within the valid range of the cosine function that will make the equation equal to zero. Therefore, the equation \((\cos x)^{99}+4 \cos x-6=0\) has no solutions.

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