Problem 17 - Algebra and Trigonometry

Question

Find all solutions of the equation. $$\cos x \sin x-2 \cos x=0$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{\pi}{2} + k\pi, \ k \in \mathbb{Z} \).

Step 1: Factor the Equation

First, factor the given equation to simplify it. The equation is \( \cos x \sin x - 2 \cos x = 0 \). Notice that \( \cos x \) is a common factor. So, factor \( \cos x \) out: \[ \cos x (\sin x - 2) = 0 \]

Step 2: Set each Factor to Zero

Use the zero product property, which states if a product of factors is zero, then at least one of the factors must be zero. Set each factor equal to zero:1. \( \cos x = 0 \)2. \( \sin x - 2 = 0 \)

Step 3: Solve \( \cos x = 0 \)

Solve for \( x \) when \( \cos x = 0 \). The cosine function equals zero at odd multiples of \( \frac{\pi}{2} \):\[ x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z} \]

Step 4: Solve \( \sin x - 2 = 0 \)

Solve for \( x \) when \( \sin x - 2 = 0 \). This implies:\[ \sin x = 2 \]Since the sine function has values only in the range \([-1, 1]\), \( \sin x = 2 \) has no real solutions.

Step 5: Combine Solutions

Combine the valid solutions from the previous steps. The first factor provided us with solutions, and the second factor gave us no solution in the real numbers. Therefore, the solutions are:\[ x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z} \]

Key Concepts

Factoring Equations

Factoring equations involves breaking down a complex equation into simpler components, which makes it easier to solve. The principle behind this is to express the original equation as a product of factors. For example, with the equation \( \cos x \sin x - 2 \cos x = 0 \), we recognize that \( \cos x \) is a common factor in both terms. By factoring out \( \cos x \), the equation simplifies to \( \cos x (\sin x - 2) = 0 \).
This step is crucial because it transforms the original expression into a format where tools like the zero product property can be applied effectively. Factoring simplifies the process of solving for variables, especially when dealing with trigonometric identities or polynomials.

Zero Product Property

The zero product property is a fundamental concept in algebra that states if a product of two factors equals zero, then at least one of the factors must be zero. In the equation \( \cos x (\sin x - 2) = 0 \), this property allows us to split the problem into simpler parts:

• \( \cos x = 0 \)
• \( \sin x - 2 = 0 \)

By setting each factor equal to zero, you essentially break the equation into more manageable sections. This method not only saves time but also clearly points to possible solutions. It's essential to use this property in scenarios where equations are factored, as it directly leads to identifying the solutions.

Cosine Function

The cosine function is one of the fundamental trigonometric functions. It describes the horizontal component of an angle within a right triangle or unit circle. The cosine of an angle \( x \) is denoted as \( \cos x \). In terms of solving equations like \( \cos x = 0 \), we focus on the specific values of angle \( x \) that make this true.
Cosine equals zero at odd multiples of \( \frac{\pi}{2} \), which means:

• \( x = \frac{\pi}{2} + k\pi \)

where \( k \) is any integer. This periodic property of the cosine function is essential in finding solutions over multiple cycles, typical in trigonometric contexts.

Sine Function

The sine function, denoted as \( \sin x \), is another key trigonometric function, representing the vertical component of an angle within a right triangle or unit circle. In our exercise, this function appears in the factor \( \sin x - 2 \).
However, the sine function has particular restrictions. Its values range only between \(-1\) and \(1\). Hence, when the equation requires solving \( \sin x = 2 \), it becomes apparent that no real solutions exist because 2 is outside of its allowable range.
Understanding the sine function's range is crucial for solving trigonometric equations, as it quickly indicates whether certain proposed solutions are feasible or not.