Problem 28 - Calculus

Question

Find a solution to the equation if possible. Give the answer in exact form and in decimal form. $$1=8 \cos (2 x+1)-3$$

Step-by-Step Solution

Answer

Exact solutions: \( x = \frac{\pi}{6} - \frac{1}{2} + k\pi \), \( x = \frac{5\pi}{6} - \frac{1}{2} + k\pi \). Decimal: 0.0236, 2.6180 (for \( k = 0 \)).

Step 1: Isolate the Trigonometric Function

Begin by isolating the cosine function on one side of the equation. Start with the given equation:\[ 1 = 8 \cos(2x + 1) - 3 \]Add 3 to both sides of the equation to isolate the cosine term on the right-hand side:\[ 4 = 8 \cos(2x + 1) \]

Step 2: Solve for Cosine by Dividing

Next, divide both sides of the equation by 8 to solve for the cosine:\[ \frac{4}{8} = \cos(2x + 1) \]Simplify the fraction:\[ \cos(2x + 1) = \frac{1}{2} \]

Step 3: Use the Inverse Cosine to Find Solutions

We know that \( \cos(\theta) = \frac{1}{2} \) corresponds to angles where the cosine value is \( \frac{1}{2} \). The principal solutions in radians are \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{5\pi}{3} \). Thus, we set:\[ 2x + 1 = \frac{\pi}{3} + 2k\pi\]and\[ 2x + 1 = \frac{5\pi}{3} + 2k\pi \]for integer values of \( k \) due to the periodicity of the cosine function.

Step 4: Solve for x

First, solve the equation \( 2x + 1 = \frac{\pi}{3} + 2k\pi\):\[ 2x = \frac{\pi}{3} - 1 + 2k\pi \]Divide every term by 2:\[ x = \frac{\pi}{6} - \frac{1}{2} + k\pi \]Next, solve the equation \( 2x + 1 = \frac{5\pi}{3} + 2k\pi \):\[ 2x = \frac{5\pi}{3} - 1 + 2k\pi \]Divide every term by 2:\[ x = \frac{5\pi}{6} - \frac{1}{2} + k\pi \]

Step 5: Convert to Decimal Form

To convert the exact form solutions into decimal forms, substitute \( \pi \approx 3.14159 \).For the solution \( x = \frac{\pi}{6} - \frac{1}{2} \) when \( k=0 \):\[ x \approx \frac{3.14159}{6} - 0.5 \approx 0.0236 \]For the solution \( x = \frac{5\pi}{6} - \frac{1}{2} \) when \( k=0 \):\[ x \approx \frac{5 \times 3.14159}{6} - 0.5 \approx 2.6180 \]

Step 6: Solution Summary

The solutions in exact form are \( x = \frac{\pi}{6} - \frac{1}{2} + k\pi \) and \( x = \frac{5\pi}{6} - \frac{1}{2} + k\pi \). The initial solutions in decimal form for \( k = 0 \) are approximately 0.0236 and 2.6180.

Key Concepts

Trigonometric Functions

Trigonometric functions are at the core of many mathematical problems involving angles and periodic phenomena. They relate the angles of a triangle to the lengths of its sides in trigonometry. The most common functions include sine, cosine, and tangent. Each of these functions has a specific role:

• **Sine** (sin) measures the ratio of the opposite side to the hypotenuse of a right-angled triangle.
• **Cosine** (cos) measures the ratio of the adjacent side to the hypotenuse.
• **Tangent** (tan) is the ratio of the sine to the cosine function, representing the opposite side over the adjacent side.

In our equation, we dealt with the cosine function, particularly focused on its values given a specific argument. Cosine values range from -1 to 1, and they follow a wave-like pattern. This property is utilized in solving trigonometric equations.

Inverse Cosine

The inverse cosine, also known as arccosine, is the function used to determine the angle whose cosine is a specified number. This function helps in solving equations where the cosine value is known, but the angle is not.To understand the inverse cosine:\(_ \arccos{(x)} \), assume you have an equation like \( \cos(\theta) = x \), where \(-1 \le x \le 1 \). The goal is to find \(\theta\), that angle whose cosine is \(x\). For our problem, \(\arccos\left(\frac{1}{2}\right)\) gave us principal angle solutions \( \frac{\pi}{3} \) and \( \frac{5\pi}{3} \) in radians. It's important to account for the periodic nature of trigonometric functions, meaning there can be infinitely many solutions if not constrained by a certain interval. This is why we include the term \(2k\pi\), where \(k\) is an integer, to represent all possible solutions.

Solving Equations

Solving trigonometric equations often involves isolating the trigonometric function and then using inverse trigonometric functions to find the angles. Let's break down the process:

• First, isolate the trigonometric term. In our example, we simplified the equation to \(\cos(2x + 1) = \frac{1}{2}\).
• Next, use the inverse cosine to find the principal solutions. We used \(\arccos\left(\frac{1}{2}\right)\) to find those starting angle values.
• Now, account for the periodic nature of the cosine function by adding \(2k\pi\) to the principal solutions, which captures all solutions along the unit circle.
• Finally, solve these equations in terms of \(x\) by manipulating algebraically to find every possible \(x\) value that satisfies the equation.

Converting the exact forms to decimal forms, while providing insight into practical measurement, can be done by substituting \(\pi\) with an approximation like 3.14159. This process helps in understanding both exact and numerical representations.