Problem 19 - Algebra and Trigonometry

Question

Find all solutions of each equation. $$ 2 \cos x+\sqrt{3}=0 $$

Step-by-Step Solution

Answer

The solution set to the equation consisting of all x that satisfy \(x = \frac{5 \pi}{6} + 2n\pi\) or \(x = \frac{7 \pi}{6} + 2n\pi\), where n denotes any integer.

Step 1: Isolate the cosine function

Subtract \(\sqrt{3}\) from both sides of the equation to isolate the cosine function. This results in: \(2 \cos x = - \sqrt{3}\)

Step 2: Solve for cosine of x

Now, divide both sides of the equation by 2: \(\cos x = - \frac{\sqrt{3}}{2}\)

Step 3: Find the reference angle

The reference angle, \(x_0\), is the acute angle within the unit circle that corresponds to \(\frac{\sqrt{3}}{2}\). We know from the unit circle that \(\cos ( \frac{\pi}{6} ) = \frac{\sqrt{3}}{2}\), so \(x_0 = \frac{\pi}{6}\). Inside calculation is done with the positive value of \(\cos x\), although \(\cos x\) is negative here.

Step 4: Find x within the correct quadrant

The cosine function is negative in the second (from \(\frac{\pi}{2}\) to \(\pi\)) and third (\(pi\) to \(\frac{3 \pi}{2}\)) quadrants of the unit circle. Thus, \(x = \pi - x_0\) or \(x = \pi + x_0\), which becomes \(x = \frac{5 \pi}{6}\) or \(x = \frac{7 \pi}{6}\).

Step 5: Accounting for All Solutions

Since cosine is periodic with period \(2\pi\), we can add \(2n\pi\) to each solution (where \(n\) is an integer) to account for all possible solutions. Therefore, the final solutions are \(x = \frac{5 \pi}{6} + 2n\pi\) or \(x = \frac{7 \pi}{6} + 2n\pi\).

Key Concepts

Unit Circle

The unit circle is an essential tool in trigonometry for understanding the behavior of trig functions. It is a circle with a radius of 1 centered at the origin of a coordinate plane.

• Every point on the unit circle represents the position of an angle in radians, originating from the positive x-axis.
• The coordinates of each point can be expressed as \( (\cos \theta, \sin \theta) \), where \( \theta \) is the angle in radians.

The unit circle is especially useful when solving trigonometric equations because it shows all possible values of sine and cosine, and their respective signs in different quadrants.
It also helps identify reference angles, which are key for finding solutions across different quadrants.

Cosine Function

The cosine function is one of the primary trigonometric functions and is frequently represented on the unit circle. It's used to calculate the horizontal coordinate of a point along the unit circle.

• The general formula for cosine is \( \cos(\theta) = \frac{Adjacent}{Hypotenuse} \) in a right triangle.
• On the unit circle, since the hypotenuse is always 1, this reduces to just the x-coordinate of a point.

The cosine graph varies between -1 and 1 as the angle changes from 0 to \( 2\pi \), periodically repeating every \( 2\pi \).
In our original exercise, we solved \( \cos x = -\frac{\sqrt{3}}{2} \), ensuring the value is negative as per the constraints of specified quadrants.

Quadrants

The coordinate plane is divided into four quadrants, which are essential for determining the signs of trigonometric functions, including cosine.

• First Quadrant: Both sine and cosine values are positive.
• Second Quadrant: Sine is positive, cosine is negative.
• Third Quadrant: Sine is negative, cosine is negative.
• Fourth Quadrant: Sine is negative, cosine is positive.

In our exercise, the solution \( \cos x = -\frac{\sqrt{3}}{2} \) led us to explore the second and third quadrants. Cosine is negative in these quadrants, alerting us to use these areas for determining the most accurate angle solutions.

Reference Angle

A reference angle is an important concept that simplifies the process of finding multiple values of an angle that share the same sine or cosine value.

• It is the smallest angle between the terminal side of the angle and the x-axis.
• The reference angle is always acute, meaning it falls between 0 and \( \frac{\pi}{2} \).

In the solution provided, we identified our reference angle first by finding the positive value \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\).
Then, we used this reference angle to determine potential angles in the relevant second and third quadrants, ensuring the correct sign for cosine is applied, thus yielding solutions \( x = \frac{5\pi}{6} \) and \( x = \frac{7\pi}{6} \).