Problem 20 - Algebra 2

Question

Solve each equation for \(0 \leq \theta<2 \pi\). $$ 3 \cos \theta=2 $$

Step-by-Step Solution

Answer

The solutions for \( \theta \) within the given domain that satisfy the equation are \( \theta = \arccos(\frac{2}{3}) \) and \( \theta = 2\pi - \arccos(\frac{2}{3}) \).

Step 1: Rewrite the equation

We'll start by isolating cosine in the equation, \( 3 \cos \theta=2.\)To do this, we'll divide both sides by 3 to solve for \(\cos \theta\),\[\cos \theta = \frac{2}{3} \].

Step 2: Use the unit circle to find \( \theta \) within domain

Now, with the value of cosine, we'll use the unit circle to find the values of \( \theta \) that fall within the specified domain of \(0 \leq \theta<2 \pi \). For a right-triangle, the adjacent side for the angle \( \theta \) would be \(2\) and the hypotenuse would be \(3\). This results in two angle positions (quadrants) on the unit circle where this triangle could occur, Quadrant I and Quadrant IV. We know that \( \cos \theta =\frac{2}{3}\) in the first quadrant, and \( \cos \theta =\frac{2}{3}\) in the fourth quadrant because the cosine function is positive in these quadrants.

Step 3: Calculate \( \theta \)

The value of \( \theta \) can be obtained by taking the inverse cosine of \( \frac{2}{3} \) on both quadrants. For quadrant I, \( \theta= \arccos(\frac{2}{3}) \).The value \( \theta \) for Quadrant IV can be found using the identity \( \theta2 = 2\pi - \theta1 \).So, \( \theta2 = 2\pi - \arccos(\frac{2}{3}) \).

Key Concepts

Unit Circle

The unit circle is a fundamental concept in trigonometry and is used to understand how the trigonometric functions behave. The unit circle is a circle with a radius of 1, centered at the origin of a coordinate plane. It helps us visualize the values of sine, cosine, and tangent for various angles.

• On the unit circle, the cosine of an angle is the x-coordinate of the point where the terminal side of the angle intersects the circle.
• The angle is measured in radians, moving counterclockwise from the positive x-axis.
• The complete rotation around the unit circle is equal to \(2\pi\) radians, which corresponds to 360 degrees.

When solving for trigonometric equations, such as finding when \(\cos \theta = \frac{2}{3}\), the unit circle helps in identifying potential angles that agree with the trigonometric equation within a specified domain, like from 0 to \(2\pi\).

Inverse Cosine

Inverse trigonometric functions help us find angles when we know the trigonometric ratio. For cosine, the inverse function is called "arccosine," denoted as \(\arccos\), allowing us to find the angle whose cosine is a known value.

• For example, if \(\cos \theta = \frac{2}{3}\), we find \(\theta\) in the equation by calculating \(\theta = \arccos(\frac{2}{3})\).
• The range of the inverse cosine function is restricted to the interval \([0, \pi]\), which allows us to find solutions in the first and second quadrants.

When solving an equation like \(3 \cos \theta = 2\), after finding \(\cos \theta = \frac{2}{3}\), using the inverse cosine helps us determine specific angles corresponding to this value in appropriate quadrants.

Quadrants in Trigonometry

Trigonometry divides a coordinate plane into four quadrants, each representing a region where the signs of trigonometric functions vary.

• Quadrant I: Both sine and cosine are positive.
• Quadrant II: Sine is positive, cosine is negative.
• Quadrant III: Both sine and cosine are negative.
• Quadrant IV: Sine is negative, cosine is positive.

Understanding which quadrants trigonometric values are positive or negative is essential in solving equations. For the equation \(\cos \theta = \frac{2}{3}\), angles that satisfy the cosine condition are located in Quadrants I and IV, where cosine is positive. Thus, once we determine \(\theta1\) using inverse cosine, we can find \(\theta2\) in Quadrant IV by subtracting \(\theta1\) from \(2\pi\).