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Chapter 4: Problem 48

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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cosine Function
The cosine function is one of the fundamental trigonometric functions encountered in mathematics. It is crucial in various fields, such as physics, engineering, and computer science. Key characteristics of the cosine function include:
  • **Domain**: The cosine function is defined for all real numbers, meaning you can input any real number into the function.
  • **Range**: Although the domain is all real numbers, the range (the set of possible output values) is limited to \([-1, 1]\). This means any angle's cosine value lies between -1 and 1, inclusive.
  • **Periodicity**: The cosine function is periodic, repeating its values every \(2\pi\) radians. This periodic nature makes it an essential function in modeling cyclical phenomena such as sound and light waves.
Cosine values are often used within equations, especially those involving oscillating or wave-like functions. For the given equation, understanding the range limitation helped determine possible solutions. Any solution must meet the criteria where the output falls within the valid range of \([-1, 1]\).
Polynomial Function
A polynomial function is a mathematical expression that involves a sum of powers in one or more variables multiplied by coefficients. The given exercise involves a polynomial in terms of \(\cos x\) expressed as \(f(\cos x) = (\cos x)^{99}+4 \cos x-6\).Essential features of polynomial functions include:
  • **Degree**: This is determined by the highest power of the variable. In this example, the degree is 99, indicating a high level of complexity due to the \(\cos x^{99}\) term.
  • **Continuity and Differentiability**: Polynomials are inherently continuous and differentiable across their entire domain. This means that \(f(\cos x)\) has no breaks or cusps, ensuring smooth curves.
  • **Zeros**: Finding zeros or solutions means identifying points where the polynomial equals zero. Here, zeros would correspond to any values of \(\cos x\) making \(f(\cos x) = 0\).
To determine solvability, the polynomial function was analyzed over the possible standard outputs of the cosine function. This context highlighted the importance of understanding both polynomial behavior and trigonometric properties.
Intermediate Value Theorem
The Intermediate Value Theorem (IVT) is a staple concept in calculus, providing insight into the presence of roots within continuous functions over a certain interval.The IVT states that if a function \(f\) is continuous over a closed interval \[a, b\]\ and takes on different signs at the endpoints, then the function must cross zero at some point within that interval.In applying the IVT to the equation \((\cos x)^{99}+4 \cos x-6=0\), the necessary checks were:
  • The function \(f(\cos x)\) is continuous, being a polynomial.
  • Evaluate at the interval endpoints \([-1, 1]\): \(f(-1) = -11\) and \(f(1) = -1\).
  • Both results are negative, indicating the function does not cross zero between \(-1\) and \(1\).
Thus, the theorem confirms no solution exists with \(\cos x\) within this range. The IVT is powerful, ensuring no 'hidden' zeros exist beyond those initially identified by endpoint checks.

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Suppose \(u\) and \(v\) are in the interval \(\left(\frac{\pi}{2}, \pi\right),\) with $$ \tan u=-2 \text { and } \tan v=-3 $$ Find exact expressions for the indicated quantities. $$ \tan (u-9 \pi) $$

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