Problem 33 - Algebra and Trigonometry

Question

Find all solutions of the equation. $$\tan \frac{x}{4}+\sqrt{3}=0$$

Step-by-Step Solution

Answer

The solutions for the equation are \( x = \frac{10\pi}{3} + 4k\pi \) or \( x = \frac{22\pi}{3} + 4k\pi \), where \( k \) is any integer.

Step 1: Isolate the Trigonometric Function

The given equation is \( \tan \frac{x}{4} + \sqrt{3} = 0 \). Begin by isolating the tangent function:\[ \tan \frac{x}{4} = -\sqrt{3} \]

Step 2: Determine the General Solution

We know that \( \tan \theta = -\sqrt{3} \) at special angles. The tangent of \( 150^{\circ} \) (which is \( \frac{5\pi}{6} \) radians) and \( 330^{\circ} \) (which is \( \frac{11\pi}{6} \)) both satisfy \( \tan \theta = -\sqrt{3} \). Thus, the general solution for \( \theta \) if \( \tan \theta = -\sqrt{3} \) is:\[ \theta = \frac{5\pi}{6} + k\pi \quad \text{or} \quad \theta = \frac{11\pi}{6} + k\pi \]where \( k \) is any integer.

Step 3: Replace the Angle with \( \frac{x}{4} \)

Substitute \( \frac{x}{4} \) back into the equation, replacing \( \theta \):\[ \frac{x}{4} = \frac{5\pi}{6} + k\pi \quad \text{or} \quad \frac{x}{4} = \frac{11\pi}{6} + k\pi \]

Step 4: Solve for \( x \)

To find \( x \), multiply both sides of each equation by 4 to solve for \( x \):1. \[ x = 4\left(\frac{5\pi}{6} + k\pi\right) = \frac{20\pi}{6} + 4k\pi = \frac{10\pi}{3} + 4k\pi \]2. \[ x = 4\left(\frac{11\pi}{6} + k\pi\right) = \frac{44\pi}{6} + 4k\pi = \frac{22\pi}{3} + 4k\pi \]

Step 5: General Solution for \( x \)

Therefore, the solutions for \( x \) are:\[ x = \frac{10\pi}{3} + 4k\pi \quad \text{or} \quad x = \frac{22\pi}{3} + 4k\pi \]where \( k \) is any integer.

Key Concepts

Tangent Function

The tangent function, often denoted as \( \tan \theta \), is a fundamental trigonometric function that relates to the ratio of the opposite side to the adjacent side in a right triangle. It's one of the three primary trigonometric functions, the other two being sine and cosine. Unlike sine and cosine, which have a range between -1 and 1, the tangent function can take on any real number value. This is because it's defined as the ratio of sine to cosine \( \left( \tan \theta = \frac{\sin \theta}{\cos \theta} \right) \).
The tangent function is periodic with a period of \( \pi \), meaning it repeats its values every \( \pi \) radians. Key properties include:

• It is undefined when \( \cos \theta = 0 \), leading to vertical asymptotes.
• It crosses the x-axis at integer multiples of \( \pi \), i.e., \( n\pi \) where \( n \) is an integer.
• Its standard pattern involves rising from negative to positive infinity over an interval from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).

Understanding these characteristics is important when solving equations involving the tangent, such as isolating \( \tan \) to find solutions.

General Solution

In trigonometry, the general solution of an equation is a way to express all possible solutions of a trigonometric function. Since the primary trigonometric functions are periodic, they repeat their values in a regular pattern which allows multiple solutions over their infinite domains. When solving equations like \( \tan \theta = -\sqrt{3} \), it's crucial to recognize this periodicity.
For the tangent function, the general solution can be outlined as:

• The solution at one known angle \( \theta_0 \), e.g., \( \frac{5\pi}{6} \) for \( \tan \theta = -\sqrt{3} \).
• Adding integer multiples of the period \( \pi \) because \( \tan(\theta + k\pi) = \tan \theta \).

Thus, the general solution for \( \tan \theta = -\sqrt{3} \) becomes \( \theta = \frac{5\pi}{6} + k\pi \) or \( \theta = \frac{11\pi}{6} + k\pi \), where \( k \) is any integer. This method ensures all possible angles that satisfy the equation are represented.

Special Angles

Special angles are specific angles with known trigonometric values, which are often used as reference points. These include \(0^\circ, 30^\circ, 45^\circ, 60^\circ,\) and \(90^\circ\), and their respective equivalents in radians. For each angle, the values of sine, cosine, and tangent are well-defined and often memorized due to their simplicity.
In the context of tangent functions, knowing the special angles where \( \tan \theta \) assumes certain values greatly simplifies solving equations. For this exercise:

• At \( 150^\circ \) (or \( \frac{5\pi}{6} \) radians), \( \tan \theta = -\sqrt{3} \).
• Similarly, at \( 330^\circ \) (or \( \frac{11\pi}{6} \) radians), \( \tan \theta = -\sqrt{3} \).

Recognizing these angles expedites finding solutions as they provide a starting point to construct general solutions using periodicity.

Integer Solutions

When dealing with trigonometric equations, integer solutions typically involve using an integer variable, often denoted as \( k \), to express solutions in a compact, general form. In a periodic function like tangent, which repeats every \( \pi \) radians, integer solutions are crucial to accounting for all possible cases.
In our solution, \( k \) represents the multiple of the period \( \pi \) that can be added to the principal solution, creating a family of solutions. Specifically:

• For \( \tan \theta = -\sqrt{3} \), any valid solution can be expressed with \( \theta = \frac{5\pi}{6} + k\pi \) or \( \theta = \frac{11\pi}{6} + k\pi \).
• This is expanded to the final form with \( x = 4\left(\frac{5\pi}{6} + k\pi\right) = \frac{10\pi}{3} + 4k\pi \) etc., by solving for \( x \).

The beauty of integer solutions lies in their ability to succinctly encode an infinite set of answers in a compact expression.