Problem 18 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation for solutions over the interval \([0,2 \pi)\) by first solving for the trigonometric finction. Do not use a calculator. $$(\cot x-\sqrt{3})(2 \sin x+\sqrt{3})=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{\pi}{6}, \frac{7\pi}{6}, \frac{4\pi}{3}, \frac{5\pi}{3}\).

Step 1: Recognize the Zero Product Property

The expression \((\cot x - \sqrt{3})(2 \sin x + \sqrt{3}) = 0\) implies that either \(\cot x - \sqrt{3} = 0\) or \(2 \sin x + \sqrt{3} = 0\). This property states that if the product of two factors equals zero, then at least one of the factors must be zero.

Step 2: Solve for \(\cot x\)

For the equation \(\cot x - \sqrt{3} = 0\), solve for \(\cot x\):\[\cot x = \sqrt{3}\]This is true for angles where the tangent is the reciprocal of \(\sqrt{3}\).

Step 3: Identify Angles for \(\cot x = \sqrt{3}\)

In the interval \([0, 2\pi)\), \(\cot x = \sqrt{3}\) at angles where the tangent is \(\frac{1}{\sqrt{3}}\), specifically at \(x = \frac{\pi}{6}\) and \(x = \frac{7\pi}{6}\).

Step 4: Solve for \(\sin x\)

For the equation \(2 \sin x + \sqrt{3} = 0\), solve for \(\sin x\):\[2 \sin x = -\sqrt{3}\]Therefore, \[\sin x = -\frac{\sqrt{3}}{2}\].

Step 5: Identify Angles for \(\sin x = -\frac{\sqrt{3}}{2}\)

In the interval \([0, 2\pi)\), \(\sin x = -\frac{\sqrt{3}}{2}\) at angles \(x = \frac{4\pi}{3}\) and \(x = \frac{5\pi}{3}\).

Step 6: Compile All Solution Angles

The solutions to the original equation, based on the individual factors, are:\(x = \frac{\pi}{6}, \frac{7\pi}{6}, \frac{4\pi}{3}, \frac{5\pi}{3}\).

Key Concepts

Zero Product Property

The Zero Product Property is a foundational concept in algebra. It states that if two or more factors multiplied together result in zero, then at least one of the factors must be zero. This idea helps simplify complex equations, making them more manageable to solve.
In the context of solving trigonometric equations, like \((\cot x - \sqrt{3})(2 \sin x + \sqrt{3}) = 0\), we apply this property directly.

• Each factor needs to be set equal to zero individually: \(\cot x - \sqrt{3} = 0\) and \(2 \sin x + \sqrt{3} = 0\).

This simplification is powerful because it breaks down the problem into smaller, solvable parts. By solving each simpler equation, we find all possible solutions to the original problem in the specified interval.

Cotangent Function

The cotangent function, \(\cot x\), is the reciprocal of the tangent function. It is defined as the ratio of the adjacent side to the opposite side in a right triangle, or mathematically,\[\cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x}.\]
Understanding this function's behavior is key in solving equations where it appears. For example, when solving \(\cot x = \sqrt{3}\), you seek angles where the tangent is \(\frac{1}{\sqrt{3}}\).

• In the interval \([0, 2\pi)\), these angles are \(x = \frac{\pi}{6}\) and \(x = \frac{7\pi}{6}\).

The cotangent function has a period of \(\pi\), meaning it repeats every \(\pi\) radians. This periodic nature is crucial for identifying possible solutions within specified intervals.

Sine Function

The sine function, denoted as \(\sin x\), is one of the primary functions in trigonometry. It represents the ratio of the length of the opposite side to the hypotenuse in a right triangle. The values of \(\sin x\) range between -1 and 1.
When solving the equation \(2 \sin x + \sqrt{3} = 0\), it needs to be simplified to find \(\sin x\):

• Solve for \(\sin x\) as \(-\frac{\sqrt{3}}{2}\).

Finding angles that satisfy this condition involves recognizing key angles on the unit circle:

• These angles within the interval \([0, 2\pi)\) are \(x = \frac{4\pi}{3}\) and \(x = \frac{5\pi}{3}\).

The sine function's periodicity of \(2\pi\) ensures that these are the correct angles for this interval.

Unit Circle

The unit circle is an essential tool in trigonometry. It is a circle with a radius of 1, centered at the origin of a coordinate plane. The unit circle links angle measures with coordinates, facilitating the calculation of trigonometric functions.
In this context, understanding sine and cotangent's behavior on the unit circle helps to identify where these functions meet specific values. For example:

• Angles like \(\frac{\pi}{6}\) and \(\frac{7\pi}{6}\) where \(\cot x = \sqrt{3}\) correlate with specific points on the circle.
• Similarly, angles \(\frac{4\pi}{3}\) and \(\frac{5\pi}{3}\) where \(\sin x = -\frac{\sqrt{3}}{2}\) also correspond to exact points.

Using the coordinates from the unit circle simplifies solving trigonometric equations by matching angle measures to their respective sine, cosine, or tangent values.