Problem 51 - Precalculus Mathematics for Calculus

Question

Solve the given equation. $$\cos \theta(2 \sin \theta+1)=0$$

Step-by-Step Solution

Answer

\( \theta = \frac{\pi}{2} + k\pi \), \( \theta = \frac{7\pi}{6} + 2k\pi \), and \( \theta = \frac{11\pi}{6} + 2k\pi \), \( k \in \mathbb{Z} \).

Step 1: Identify Factors

The equation is given as \( \cos \theta (2 \sin \theta + 1) = 0 \). Identify that the equation represents two separate factors: \( \cos \theta = 0 \) and \( 2 \sin \theta + 1 = 0 \). Each factor can be set to zero to find solutions for \( \theta \).

Step 2: Solve for \( \cos \theta = 0 \)

The equation \( \cos \theta = 0 \) holds true when \( \theta \) is of the form \( \theta = \frac{\pi}{2} + k\pi \), where \( k \) is an integer. These are the angles where the cosine function crosses the x-axis.

Step 3: Solve for \( 2 \sin \theta + 1 = 0 \)

Simplify to find \( \sin \theta \). Subtract 1 from both sides to get \( 2 \sin \theta = -1 \), and then divide by 2 to find \( \sin \theta = -\frac{1}{2} \).

Step 4: Find angles for \( \sin \theta = -\frac{1}{2} \)

The sine function equals \(-\frac{1}{2}\) at angles \( \theta = \frac{7\pi}{6} + 2k\pi \) and \( \theta = \frac{11\pi}{6} + 2k\pi \), where \( k \) is any integer. These are the angles where the sine function has the corresponding value in the unit circle.

Key Concepts

Solving Trigonometric Equations

Understanding how to solve trigonometric equations is a key skill in mathematics. A trigonometric equation is one that involves sine, cosine, tangent, or any other trigonometric function. To solve these, you typically want to isolate each trigonometric function as much as possible. In our example, the equation \( \cos \theta(2 \sin \theta + 1) = 0 \) involves both cosine and sine functions. Breaking it into factors helps simplify the process. We treat each part of the product as a separate equation:

• \( \cos \theta = 0 \)
• \( 2 \sin \theta + 1 = 0 \)

Setting each of these factors to zero allows us to find possible solutions for \( \theta \), leveraging the periodic nature of trigonometric functions.
Learning how to efficiently address each component separately helps else us find a comprehensive solution. By setting the products in the equation to zero, you can solve them as linear trigonometric problems. This method also reinforces the roots of trigonometric functions related to the unit circle.

Cosine Function

The cosine function, represented as \( \cos \theta \), is essential in trigonometry. It tells us the horizontal position of a point on the unit circle, starting from the angle \( \theta \) measured from the positive x-axis.For the equation \( \cos \theta = 0 \), we need to determine when the cosine value crosses the x-axis. This occurs at \( \theta = \frac{\pi}{2} + k\pi \), where \( k \) is an integer. It happens because \( \cos \theta \) equals zero at these angles, indicating vertical positioning.Understanding these fixed points can be particularly useful and simplifies solving equations. Remember:

• \( \theta = \frac{\pi}{2} + k\pi \) are crucial points where cosine zeroes out.
• Each integer \( k \) corresponds to another solution, reflecting the periodic nature of cosine.

With these key insights, you can easily solve equations involving cosine zero-crossings.

Sine Function

Similar to cosine, the sine function, noted as \( \sin \theta \), measures the vertical component of a point on the unit circle. For our equation, the task is to solve \( 2 \sin \theta + 1 = 0 \).Start by isolating \( \sin \theta \):

• Subtract 1 from both sides: \( 2 \sin \theta = -1 \)
• Divide by 2: \( \sin \theta = -\frac{1}{2} \)

Now we need to determine where \( \sin \theta = -\frac{1}{2} \) on the unit circle. The sine function takes this value at specific points which are:

• \( \theta = \frac{7\pi}{6} + 2k\pi \)
• \( \theta = \frac{11\pi}{6} + 2k\pi \)

These angles correspond to standard positions in the unit circle where sine equals \(-\frac{1}{2}\). Recognizing these angles come from knowing the unit circle and the periodicity of the sine function. With this, any trigonometric equation involving sine becomes much simpler to handle.