Problem 4 - Precalculus

Question

In Exercises \(1-18,\) find all of the exact solutions of the equation and then list those solutions which are in the interval \([0,2 \pi)\). $$ \tan (6 x)=1 $$

Step-by-Step Solution

Answer

The exact solutions in [0, 2π) are π/24, 5π/24, 3π/8, 13π/24, 17π/24, 7π/8, 25π/24, 29π/24, 11π/8, 37π/24, 41π/24, 15π/8.

Step 1: Understand the Tangent Equation

The equation given is \( \tan(6x) = 1 \). Recall that the tangent function is periodic with a period of \( \pi \). Hence, when \( \tan(\theta) = 1 \), \( \theta \) could be \( \frac{\pi}{4} \) plus any multiple of \( \pi \).

Step 2: Find General Solutions

Since \( \tan(\theta) = 1 \) leads to \( \theta = \frac{\pi}{4} + k\pi \) (where \( k \) is an integer), equate \( 6x \) to this expression: \( 6x = \frac{\pi}{4} + k\pi \).

Step 3: Solve for x

Divide both sides of the equation \( 6x = \frac{\pi}{4} + k\pi \) by 6 to solve for \( x \):\[x = \frac{1}{6} \left( \frac{\pi}{4} + k\pi \right) = \frac{\pi}{24} + \frac{k\pi}{6}\]

Step 4: List Solutions in \([0, 2\pi)\)

Find values of \( k \) such that \( x = \frac{\pi}{24} + \frac{k\pi}{6} \) remains within the interval \([0, 2\pi)\):- For \( k = 0 \), \( x = \frac{\pi}{24} \).- For \( k = 1 \), \( x = \frac{5\pi}{24} \).- For \( k = 2 \), \( x = \frac{9\pi}{24} = \frac{3\pi}{8} \).- For \( k = 3 \), \( x = \frac{13\pi}{24} \).- For \( k = 4 \), \( x = \frac{17\pi}{24} \).- For \( k = 5 \), \( x = \frac{21\pi}{24} = \frac{7\pi}{8} \).- For \( k = 6 \), \( x = \frac{25\pi}{24} \).- For \( k = 7 \), \( x = \frac{29\pi}{24} \).- For \( k = 8 \), \( x = \frac{33\pi}{24} = \frac{11\pi}{8} \).- For \( k = 9 \), \( x = \frac{37\pi}{24} \).- For \( k = 10 \), \( x = \frac{41\pi}{24} \).- For \( k = 11 \), \( x = \frac{45\pi}{24} = \frac{15\pi}{8} \).No further solutions exist as they exceed \( 2\pi \).

Step 5: Final Answer for \([0, 2\pi)\)

The solutions in the interval \([0, 2\pi)\) are:- \( \frac{\pi}{24} \)- \( \frac{5\pi}{24} \)- \( \frac{3\pi}{8} \)- \( \frac{13\pi}{24} \)- \( \frac{17\pi}{24} \)- \( \frac{7\pi}{8} \)- \( \frac{25\pi}{24} \)- \( \frac{29\pi}{24} \)- \( \frac{11\pi}{8} \)- \( \frac{37\pi}{24} \)- \( \frac{41\pi}{24} \)- \( \frac{15\pi}{8} \)

Key Concepts

Periodic Functions

Tangent functions are an excellent example of periodic functions. A periodic function is one that repeats its values at regular intervals, known as the period. For the tangent function, that period is \( \pi \). This means that \( \tan(\theta) = \tan(\theta + n\pi) \), where \( n \) is any integer. This repetitive nature is crucial when solving equations involving tangent, as it allows you to account for all possible solutions by simply finding one particular solution and adding multiples of the period.

Angle Measurement in Radians

When we work with trigonometric functions and equations, angles are usually measured in radians. This measurement is based on the radius of a circle. One full rotation around a circle is \( 2\pi \) radians, equivalent to 360 degrees. Hence, \( \pi \) radians is half a rotation, or 180 degrees, and \( \frac{\pi}{2} \) is a quarter turn, or 90 degrees. Using radians makes calculations in trigonometry more convenient and is the standard in mathematical contexts. For the equation \( \tan(6x) = 1 \), interpreting the angle \( \theta \) in radians helps easily identify solutions in terms of multiples of \( \pi \).

Trigonometric Equations

Trigonometric equations involve trigonometric functions such as sine, cosine, and tangent. These equations require specific strategies for solutions because of their periodic nature. Returning to our equation, \( \tan(6x) = 1 \), we use the property at the angle \( \theta = \frac{\pi}{4} \) (where tangent equals 1) to find solutions. By acknowledging that \( \tan \) repeats every \( \pi \) units, you can express any angle that gives a tangent of 1 as \( \theta = \frac{\pi}{4} + k\pi \). Here, \( k \) is an integer representing how many full periods you've added.

Solutions of Equations

After establishing a general form for the angles, the next step requires solving for \( x \). This often means isolating the variable you need, which in our case is \( x \). From our angle equation \( 6x = \frac{\pi}{4} + k\pi \), solving for \( x \) involves dividing every term by 6:

• \( x = \frac{\pi}{24} + \frac{k\pi}{6} \)

By plugging different integer values for \( k \), you generate specific solutions for \( x \) that are within a given interval. Here, the interval is \([0, 2\pi)\), meaning valid \( x \) values must stay within these bounds. For each \( k \), calculate \( \frac{\pi}{24} + \frac{k\pi}{6} \) to check if the result is still within \([0, 2\pi)\). This process compiles all possible solutions within the desired range.