Problem 41 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find all solutions of the equation in the interval \([0,2 \pi)\) algebraically. Use the table feature of a graphing utility to check your answers numerically. $$\cos ^{3} x=\cos x$$

Step-by-Step Solution

Answer

The solutions to the given equation are \(x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{2}\).

Step 1: Rewrite the Equation

The given trigonometric equation is \(\cos^3x = \cos x\). This can be rewritten as \(\cos x * (\cos^2x - 1) = 0\) by factoring out \(\cos x\).

Step 2: Substitute Pythagorean Identity

Then, we substitute \(\cos^2x\) with \(1 - \sin^2x\) using the Pythagorean identity. This leads to \(\cos x * (1 - \sin^2x - 1) = Cos x (1 - 2 \sin^2x) = 0\).

Step 3: Find the solutions

From the equality \(\cos x = 0\), we get \(x = \frac{1\pi}{2}, \frac{3\pi}{2}\). From the equality \(1 - 2 \sin^2x = 0\), we solve for \(\sin^2x\), leading to \(\sin^2x = \frac{1}{2}\). Again, we solve for \(x\) which gives \(x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\).

Step 4: Check solutions in the interval

Verify these solutions are within our desired interval \([0, 2\pi)\). Luckily, all of these solutions lie between 0 and \(2\pi\). Thus, are all valid solutions of the given equation.

Key Concepts

Factoring trigonometric expressions

Factoring trigonometric expressions is a useful technique when solving equations involving trigonometric functions. By factoring, we break down complex expressions into simpler multiplicative components. In the given problem, the equation \(\cos^3x = \cos x\) was factored into \(\cos x(\cos^2x - 1) = 0\). This was achieved by factoring out \(\cos x\) from both sides. It's similar to simplifying algebraic expressions like \(x^3 = x\) into \(x(x^2 - 1) = 0\).

• The key to factoring trigonometric expressions is to look for common factors or use identities.
• After factoring, the problem often reduces to finding solutions for each component separately.
• Remember to set each factor equal to zero to find the values of \(x\).

Exploring the expression "\(\cos x(\cos^2x - 1)\)," we recognize that we have created a product of two expressions. Solving each such expression individually can greatly simplify the process.

Pythagorean identity

The Pythagorean identity is among the most fundamental identities in trigonometry. It states that \(\sin^2x + \cos^2x = 1\). This identity allows you to express sine in terms of cosine and vice versa. In the problem, substituting \(\cos^2x\) with \(1 - \sin^2x\) simplifies the factoring process even further. After replacing \(\cos^2x\) in \(\cos x(\cos^2x - 1) = 0\), we obtain \(\cos x (1 - \sin^2x - 1)\), which simplifies to \(\cos x(1 - 2\sin^2x) = 0\).

• Use the identity: It helps simplify expressions or change the form that makes solving easier.
• Convert the trigonometric function as needed using \(1 - \sin^2 x = \cos^2 x\) or \(1 - \cos^2 x = \sin^2 x\).
• Substitution can lead to new equations that are easier to solve algebraically.

In practice, remember that using this identity is often necessary when no other solution seems apparent, effectively bridging the gap between different trigonometric functions.

Interval solutions

Finding solutions within a specified interval, such as \([0, 2\pi)\), is critical in trigonometric equations. It ensures that we only consider solutions that satisfy specific conditions. In the example given, the task was to find solutions to the equation within \([0, 2\pi)\). Once the equation \(\cos x(1 - 2\sin^2x) = 0\) is fully solved, potential solutions include values for \(x\) at \(\frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}\), and \(\frac{7\pi}{4}\). The reason we focus on the interval is as follows:

• Trigonometric functions are periodic, which means:
  • They repeat their values over specific intervals.
• By restricting solutions to \([0, 2\pi)\), we cover one complete cycle of the cosine and sine functions.
• Selecting the correct interval helps avoid extraneous solutions from multiple cycles.

It's important to check each solution to ensure it's within the defined interval, confirming that you're resolving the equation over the right range.