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Chapter 2: Problem 27

Solve for \(x: \cos ^{-1} x+\cos ^{-1} x^{2}=0\)

Short Answer

Expert verified
The solution to the equation is \(x = 1\).

Step by step solution

01

Analysis of the ranges of the inverse cosine function

Since the range of \(\cos^{-1}x\) is \([0, \pi]\), the given equation implies that \(\cos^{-1} x\) and \(\cos^{-1} x^2\) are both zero. Therefore, the corresponding arguments \(x\) and \(x^2\) must be 1 because the cosine of 0 is 1.
02

Solve for \(x\)

Setting both arguments to 1 gives us two equations: \(x = 1\) and \(x^2 = 1\). The second equation gives two solutions: \(x = 1\) and \(x= -1\). Therefore, the total solutions for given equation are \(x = 1\) and \(x = -1\). Note \(x = 1\) is found in both equations, so we have two distinct solutions in total. Check our answer by plugging these value back into the original equation to ensure they hold true.
03

Cross-Check the solutions

For \(x = 1\), the equation becomes \(\cos^{-1}(1) + \cos^{-1}(1) = 0 + 0 = 0\). For \(x = -1\), \(\cos^{-1}(-1) + \cos^{-1}((-1)^2) = \pi + 0 = \pi\). The second one is not true, therefore \(x = -1\) is an extraneous solution. Only \(x = 1\) is correct.

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Key Concepts

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

Range of Inverse Cosine Function
The inverse cosine function, often denoted as \( \cos^{-1}(x) \) or \( \text{arccos}(x) \), is a vital concept in trigonometry. It allows us to determine the angle whose cosine is a given number. The range of this function is
  • [0, \( \pi \)]: This means that the angle we determine from the inverse cosine function will always fall between 0 and \( \pi \) radians inclusive.
  • This range corresponds to angles in the first and second quadrants of the unit circle.
In the given exercise, understanding that both \( \cos^{-1}(x) \) and \( \cos^{-1}(x^2) \) need to equal 0 for their sum to be 0 was key. Since \( \cos^{-1}(1) = 0 \), it implies that both \( x \) and \( x^2 \) must equal 1.
Solving Trigonometric Equations
Solving trigonometric equations involves finding all values of the variable that make the equation true. Let's break down the solution process:
  • Identify the trigonometric functions involved: In this case, the inverse cosine functions \( \cos^{-1}(x) \) and \( \cos^{-1}(x^2) \).
  • Set the function inside the inverse cosine to equal the angle that yields the desired cosine value:
    • \( x = 1 \) because \( \cos 0 = 1 \).
    • \( x^2 = 1 \) which simplifies to \( x = 1 \) or \( x = -1 \).
This approach ensures we consider all potential solutions and check them against the original equation to find valid ones.
Extraneous Solutions
Extraneous solutions are values obtained during the solving process that do not satisfy the original equation. They often appear in more complex algebraic manipulations, especially when dealing with inverses or square roots.
  • When solving \( x^2 = 1 \), we mathematically derive solutions \( x = 1 \) and \( x = -1 \).
  • However, upon substituting \( x = -1 \) back into the original equation, the equation does not hold.
  • The equation becomes \( \cos^{-1}(-1) + \cos^{-1}(1) = \pi + 0 \), which equals \( \pi \), not 0 as required.
Thus, \( x = -1 \) is considered an extraneous solution, while \( x = 1 \) remains the valid and correct solution to the original problem.

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