Problem 31 - Precalculus Essentials

Question

Involve equations with multiple angles. Solve each equation on the interval \([0,2 \pi)\) $$\tan \frac{x}{2}=\sqrt{3}$$

Step-by-Step Solution

Answer

From the analysis, solving and review of the possible solutions, it can be seen that x = 2n\pi + \frac{2\pi}{3} provides viable solutions for x in the given interval. Specifically, using n as 0 and 1 will give x = \frac{2\pi}{3} and x = \frac{8\pi}{3}, respectively. However, x = 2n\pi + \frac{ 8 \pi }{3} does not provide any solutions for x in the given interval with n being an integer.

Step 1: Resolving the function

Use the special triangle and the property of the function \(\tan \Theta\). From the special triangle, it is known that \(\tan \Theta = \sqrt{3}\) for \(\Theta = \frac{ \pi }{3}\) and \(\Theta = \frac{ 4 \pi }{3}\). However, since \(\tan()\) is a periodic function with a period of \(\pi\), \(\tan \Theta = \sqrt{3}\) for \(\Theta = n\pi + \frac{\pi}{3}\) and \(\Theta = n\pi + \frac{4 \pi }{3}\), where n is an integer.

Step 2: Equating the values of Theta

In this case, \(\Theta = \frac{ x }{2}\), so it can be said that \(\frac{x }{2}\) = \( n\pi + \frac{ \pi }{3}\) and \(\frac{x }{2}\) = \(n\pi + \frac{ 4 \pi }{3}\).

Step 3: Solving for x

Multiply both sides of the equations by 2 to isolate x. This gives x =\(2n\pi + \frac{ 2 \pi }{3}\) and x = \(2n\pi + \frac{ 8 \pi }{3}\). With these results, the challenge is to find the right values of n such that x falls in the interval \([0,2 \pi)\), or \([0, 2 \times 3.14)\).

Step 4: Further analysis on the result

Let us consider \(x = 2n\pi + \frac{ 2 \pi }{3}\). For x to be within \([0,2 \pi)\), the value of \(2n\pi\) must be equal to or greater than zero and less than \(2 \times 3.14 - \frac{ 2 \pi }{3}\). Hence, n should be selected as 0 and 1 for this formula. Now, when considering \(x = 2n\pi + \frac{ 8 \pi }{3}\), for x to lie within the interval, the value of \(2n\pi\) must be greater than zero and less than \(2 \times 3.14 - \frac{ 8 \pi }{3}\). For this formula, since \(2n\pi\) should be less negative than \(- \frac{ 8 \pi }{3}\), there is no suitable integer for n.

Key Concepts

Tangent Function

The tangent function, denoted as \( \tan(x) \), is an important trigonometric function defined as the ratio of the sine and cosine functions: \( \tan(x) = \frac{\sin(x)}{\cos(x)} \). This means it is undefined for angles where \( \cos(x) = 0 \), because dividing by zero is not possible. The tangent function is well known for its unique characteristic of having a vertical asymptote at every odd multiple of \( \frac{\pi}{2} \), where the cosine is zero.
Imagine a right triangle: the tangent of an angle is the length of the opposite side divided by the length of the adjacent side. Graphically, \( \tan(x) \) shows repeating patterns because it climbs upwards to infinity and then suddenly drops to negative infinity, repeating this behavior each period.

• Asymptotic Behavior: The function climbs toward infinity near odd multiples of \( \frac{\pi}{2} \).
• Cyclic Nature: It repeats its behavior every \( \pi \) radians (180 degrees).

Understanding the behavior of the tangent function can be particularly useful in solving trigonometric equations, as it helps to anticipate where the function will repeat its output values.

Angle Resolution

Angle resolution in trigonometric equations is the process of identifying angles that satisfy a given condition within a specified interval. When resolving angles for the tangent function, we try finding all possible angles \( \Theta \) that satisfy an equation, taking into account its periodic nature.
For the equation \( \tan(\theta) = \sqrt{3} \), we look for angles for which the tangent equals \( \sqrt{3} \) using knowledge of special triangles. A special triangle with a \( \tan(\theta) = \sqrt{3} \) indicates angles such as \( \frac{\pi}{3} \) (or 60 degrees) and \( \frac{4\pi}{3} \) (or 240 degrees).

• Determine Special Angles: Identify angles from the unit circle or known triangles that satisfy the equation.
• General Solution: Write the general solution for all angles by using the fact that tangent repeats every \( \pi \) radians.

In our example, solving \( \tan(\frac{x}{2}) = \sqrt{3} \) involves setting \( \frac{x}{2} = \frac{\pi}{3} \) and \( \frac{4\pi}{3} \), and then multiplying every valid solution by 2 to "solve for \( x \)." This reasoning ultimately allows us to find all valid solutions in the desired interval.

Periodicity

Periodicity in trigonometric functions refers to their repeating behavior over regular intervals. For the tangent function, this behavior is especially prominent. The tangent of an angle repeats every \( \pi \) radians, making it a crucial factor when solving equations involving \( \tan(x) \).
Let's break down why periodicity matters:

• Repetition: If \( \tan(x) = \sqrt{3} \), it holds for \( x = \frac{\pi}{3} \) and again at \( x = \frac{\pi}{3} + n\pi \) for any integer \( n \).
• Interval Constraints: In any given interval, like \([0, 2\pi)\), periodicity lets us find all possible solutions by adding multiples of \( \pi \) to any known solution, without exceeding the bounds.

When finding arguments such as \( x = 2n\pi + \frac{2\pi}{3} \) for an interval \([0, 2\pi)\), periodicity dictates the need to calculate different \( n \) to determine feasible solutions. This ensures that all solutions respect both the equation and the interval constraints.