Problem 21 - Algebra and Trigonometry

Question

\(17-34\) . An equation is given. (a) Find all solutions of the equation. (b) Find the solutions in the interval \([0,2 \pi) .\) $$ \sqrt{3} \tan 3 \theta+1=0 $$

Step-by-Step Solution

Answer

Solutions in \([0, 2\pi)\) are \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}, \frac{13\pi}{6}\).

Step 1: Isolate the trigonometric function

Start by isolating the tangent function in the given equation. The original equation is \( \sqrt{3} \tan 3\theta + 1 = 0 \). To isolate \( \tan 3\theta \), subtract 1 from both sides and then divide by \( \sqrt{3} \): \[ \tan 3\theta = -\frac{1}{\sqrt{3}} \].

Step 2: Solve the equation for the general solution

Now, determine the values of \( \theta \) where \( \tan \theta = -\frac{1}{\sqrt{3}} \). For tangent, \( \theta = \arctan(-\frac{1}{\sqrt{3}}) \) corresponds to specific angles on the unit circle. These are \( \theta = -\frac{\pi}{6} + n\pi \) where \( n \) is an integer, considering the periodicity of tangent.

Step 3: Adjust for the multiple angle

Since the original equation is in terms of \( 3\theta \), we replace \( \theta \) with \( 3\theta \): \[ 3\theta = -\frac{\pi}{6} + n\pi \].

Step 4: Solve for \( \theta \)

Divide each term in the equation by 3 to solve for \( \theta \): \[ \theta = -\frac{\pi}{18} + \frac{n\pi}{3} \]. This gives us the general solution for \( \theta \).

Step 5: Find solutions in the interval \( [0, 2\pi) \)

To find specific solutions in the interval \([0, 2\pi)\), substitute different integer values for \( n \) in \( \theta = -\frac{\pi}{18} + \frac{n\pi}{3} \):- For \( n = 1 \): \( \theta = \frac{\pi}{6} \)- For \( n = 2 \): \( \theta = \frac{5\pi}{6} \)- For \( n = 3 \): \( \theta = \frac{9\pi}{6} = \frac{3\pi}{2} \) - For \( n = 4 \): \( \theta = \frac{13\pi}{6} \)Each of these values must fall within \([0, 2\pi)\). Any \( n \) resulting in \( \theta \) greater than \( 2\pi \) is not valid within this interval.

Key Concepts

tangent function

The tangent function, denoted as \( \tan \theta \), is one of the fundamental trigonometric functions. It relates the angle \( \theta \) in a right-angled triangle to the ratio of the opposite side over the adjacent side. However, when we look at the unit circle, this function can be extended to all real angles.

• Key Properties: The tangent function is periodic with a period of \( \pi \).
• It has asymptotes (points where the function is undefined) at angles where the cosine equals zero, such as \( \frac{\pi}{2} + n\pi \).
• The range of the tangent function is all real numbers because it can take any value from negative infinity to positive infinity.

For the equation \( \sqrt{3} \tan 3\theta + 1 = 0 \), we initially isolate the tangent to find \( \tan 3\theta = -\frac{1}{\sqrt{3}} \). This means we are looking at angles where \( \theta \) gives a tangent value of \(-\frac{1}{\sqrt{3}}\).

unit circle

The unit circle is a circle with a radius of one unit, centered at the origin of a coordinate plane. It is a crucial concept for understanding the trigonometric functions and their relationships.

• Significance of the Unit Circle: Every point on the unit circle corresponds to an angle \( \theta \) measured from the positive x-axis and makes the formation of a right triangle possible with the
  • horizontal leg of length \( \cos \theta \)
  • vertical leg of length \( \sin \theta \)
• These coordinates \((\cos \theta, \sin \theta)\) determine the sine and cosine values of \( \theta \).
• For tangent, \( \tan \theta \) is calculated as the slope of the line from the origin to the point, or \( \frac{\sin \theta}{\cos \theta} \).

To find \( \tan 3\theta = -\frac{1}{\sqrt{3}} \) on the unit circle, locate the angles where this ratio occurs. Here, the angle is \(-\frac{\pi}{6} + n\pi \), adjusting for the periodicity and specific intervals.

periodicity of trigonometric functions

Trigonometric functions have repeating patterns, known as periodicity. Understanding this is vital for solving equations that involve these functions.

• Tangent Periodicity: The periodicity of the tangent function is \( \pi \), which means every \( \pi \) units, the tangent values repeat.
• For equations involving tangent, this implies any solution \( \theta \) will repeat an angle plus \( n\pi \) for any integer \( n \).

Allowing for this periodic nature, we can express the general solution for the equation in step 2 as \( \theta = -\frac{\pi}{6} + n\pi \). When modifying these solutions for the specific problem \( \tan 3\theta = -\frac{1}{\sqrt{3}} \), it is crucial to consider \( 3\theta \) due to the multiplication factor.

general solution

Finding a general solution involves identifying all potential solutions for a trigonometric equation by considering its periodic nature.

• From a mathematical perspective, once isolated, these functions yield solutions which include all periodic repeats.
• For our equation, after isolation, we find the solution \( \theta = -\frac{\pi}{18} + \frac{n\pi}{3} \).
• This solution represents all possible values of \( \theta \) satisfying the requirement of the tangent value being \(-\frac{1}{\sqrt{3}}\).

Each \( n \) represents a complete period's increment where the solution "repeats" its value within the context of multiplication and division applied in the equation.

interval solutions

Interval solutions are particular solutions to the trigonometric equation that lie within a specified range or interval.

• For example, our interval is \([0, 2\pi)\).
• We achieve this by substituting different values of \( n \) into \( \theta = -\frac{\pi}{18} + \frac{n\pi}{3} \), where we only choose values of \( \theta \) falling within this range.
• In this scenario:
  • For \( n = 1 \): the solution is \( \theta = \frac{\pi}{6} \)
  • For \( n = 2 \): \( \theta = \frac{5\pi}{6} \)
  • For \( n = 3 \): \( \theta = \frac{3\pi}{2} \)
  • For \( n = 4 \): \( \theta = \frac{13\pi}{6} \)

This indicates that each substitution's result should make sure that the outcome remains within \([0, 2\pi)\) to be considered a valid solution within the interval.