Problem 74 - Precalculus Essentials

Question

Use an identity to solve each equation on the interval \([0,2 \pi)\) $$\cos 2 x+\cos x+1=0$$

Step-by-Step Solution

Answer

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

Step 1: Application of Double Angle Identity

Use the double angle identity \(\cos 2x = 2\cos^2x - 1\) to replace \(\cos 2x\) in the equation. This leads to the equation \(2\cos^2 x -1 + \cos x +1 =0\), which simplifies to \(2\cos^2x + \cos x = 0\).

Step 2: Factorising the Equation

The quadratic equation from step 1 can be factored as \(\cos x (2\cos x + 1) = 0\).

Step 3: Solving for \(\cos x\)

Setting each factor equal to zero gives \(\cos x = 0\) and \(2\cos x + 1 = 0\). Solving these equations gives \(x = \frac{\pi}{2}, \frac{3\pi}{2}\) for \(\cos x = 0\) and \(x = \frac{2\pi}{3}, \frac{4\pi}{3}\) for \(2\cos x + 1 = 0\).

Step 4: Verifying the Results

Verify that all solutions are in the interval [0, 2\(\pi\)) and substitute them back into the original equation to verify they are correct.

Key Concepts

Double Angle Identities

Double angle identities in trigonometry are incredibly useful tools for simplifying expressions and solving equations. One common identity is the cosine double angle identity: \( \cos 2x = 2\cos^2x - 1 \). This identity expresses \( \cos 2x \) in terms of \( \cos x \), allowing us to replace more complex expressions with simpler ones. This particular identity is useful because it transforms a problem into a possibly solvable quadratic form.

• To use this identity, substitute \( \cos 2x \) with \( 2\cos^2x - 1 \) in the equation.
• After substitution, a trigonometric equation is transformed into a quadratic form, making it easier to solve.

Applying this identity helps in a multitude of scenarios where trigonometric equations play a part, such as the equation \( \cos 2x + \cos x + 1 = 0 \) seen in our exercise. By performing the substitution, the equation \( \cos 2x + 1 = -\cos x \) simplifies directly to \( 2\cos^2x + \cos x = 0 \). This sets up the equation for easier manipulation and solution.

Factoring Quadratic Equations

Factoring quadratic equations is a fundamental algebraic technique used to eliminate variables and find solutions. When you have a quadratic equation in the form \( ax^2 + bx + c = 0 \), the goal is to express it as a product of its factors.

• For the quadratic \( 2\cos^2x + \cos x = 0 \), we can factor out the common factor \( \cos x \).
• This gives us: \( \cos x (2\cos x + 1) = 0 \).
• Setting each factor equal to zero gives you equations that are relatively simple to solve simultaneously: \( \cos x = 0 \) and \( 2\cos x + 1 = 0 \).

This method is effective because any value of \( x \) that solves each equation also solves the original equation. Factorization is a reliable technique that leverages the zero-product property, which asserts that if a product of factors equals zero, then at least one of the factors must be zero.

Intervals in Trigonometry

Understanding intervals in trigonometry is essential for determining valid solutions to equations. The interval \([0, 2\pi)\) signifies that we are seeking angles \( x \) measured in radians within the full range of a circle but not including \( 2\pi \). This notation is crucial when calculating valid solutions, ensuring they fall within a designated range.

• The interval \([0, 2\pi)\) includes all possible angles one might encounter on a standard unit circle.
• When solving equations like \( \cos x = 0 \) or \( 2\cos x + 1 = 0 \), solutions must be checked to ensure they are within this interval.
• For instance, the solutions \( x = \frac{\pi}{2}, \frac{3\pi}{2} \) satisfy \( \cos x = 0 \), while \( x = \frac{2\pi}{3}, \frac{4\pi}{3} \) satisfy \( 2\cos x + 1 = 0 \), all of which lie within \([0, 2\pi)\).

Intervals ensure that solutions are practical and applicable to real-world scenarios or designated mathematical problems, forming a crucial component of solving trigonometric equations.