Problem 63 - Precalculus: Functions and Graphs

Question

Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. \(3 \cos 2 x-7 \cos x+5=0\) \([0,2 \pi)\)

Step-by-Step Solution

Answer

In the interval, the solution is \( x = 0 \).

Step 1: Use Trigonometric Identities

This problem involves a quadratic trigonometric equation. Start by substituting a trigonometric identity if necessary. In this case, note that the equation is already in terms of \( \cos x \). Hence, we may directly proceed to solve: 3\( \cos^2x - 7\cos x + 5 = 0 \).

Step 2: Substitute Variables

Introduce a substitution to simplify the equation. Let \( y = \cos x \), which transforms the equation into a quadratic equation: 3\( y^2 - 7y + 5 = 0 \).

Step 3: Solve the Quadratic Equation

Apply the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -7 \), and \( c = 5 \). Calculate the discriminant: \( b^2 - 4ac = (-7)^2 - 4(3)(5) = 49 - 60 = -11 \). Since the discriminant is negative, there are no real solutions, suggesting we reinterpret \( y \) such that it can not use the standard quadratic formula. Instead, factor directly or by completing the square.

Step 4: Factor Quadratic Equation

Directly factor the quadratic \( 3y^2 - 7y + 5 = 0 \) by using possible trial and error or factorization techniques. The factorization is: \((3y - 5)(y - 1) = 0\). Therefore, the solutions by factoring are: \( y = \frac{5}{3} \) and \( y = 1 \).

Step 5: Convert Back to Trigonometric Terms

Since \( y = \cos x \), the solutions for \( \cos x = \frac{5}{3} \) are not valid (as the cosine function value must be between -1 and 1). So, we only evaluate \( \cos x = 1 \). This gives the solution \( x = 0 \) in the interval \([0, 2\pi)\).

Key Concepts

Trigonometric Identities

Understanding trigonometric identities is crucial when dealing with trigonometric equations. These identities are equations that involve trigonometric functions and are universally true for all angles. They help simplify complex trigonometric expressions and equations.
In this exercise, trigonometric identities are used to solve the equation \(3 \cos 2x - 7 \cos x + 5 = 0\).

• By noting that the equation is composed of one direct trigonometric function, specifically \(\cos x\), an identity substitution wasn't necessary.
• When rewriting the equation, you consider that \(\cos 2x\) could relate to \(\cos^2 x\) and \(\sin^2 x\), however, direct substitution was not needed due to the equation's simplicity.

A deeper understanding and familiarity with these identities can help you recognize patterns and quickly decide how to simplify or solve trigonometric equations. Keeping these identities handy is beneficial until you commit them to memory.

Quadratic Equations

Quadratic equations play an essential role when solving problems involving inverse trigonometric functions. A standard quadratic equation takes the form \(ax^2 + bx + c = 0\).
This exercise required converting a trigonometric equation to a quadratic form by letting \(y = \cos x\):

• This transformed the original equation into the quadratic equation \(3y^2 - 7y + 5 = 0\).
• Solving this quadratic equation usually involves factoring or using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• However, note that not all quadratic equations will have real roots, as indicated by the discriminant \(b^2 - 4ac\). If negative, it suggests complex solutions.

In this exercise, the discriminant was negative. Instead, direct factoring was practical in finding usable solutions. Understanding quadratic equations is key to transitioning from algebraic manipulation to interpreting trigonometric solutions.

Cosine Function

The cosine function is one of the primary trigonometric functions. It is defined as the adjacent side over the hypotenuse in a right triangle. Cosine values always range between -1 and 1. This exercise involved determining valid cosine values using the quadratic equation obtained.

• Once you find values for \(y\) in a trigonometric equation, they are interpreted as cosine values \(\cos x\). These values must satisfy the range restrictions of the cosine function.
• In this exercise, one potential solution was \(y = \frac{5}{3}\), but this was rejected since it falls outside the valid range.
• Ultimately, \(y = 1\) was the only valid solution, resulting in \(\cos x = 1\).

This solution reveals \(x = 0\) is the angle yielding a cosine of 1 within the interval \([0, 2\pi)\). Understanding the range and properties of the cosine function is vital for accurately solving trigonometric equations.