Problem 32 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-38: Find all solutions of the equation. $$ (2 \sin u-1)(\cos u-\sqrt{2})=0 $$

Step-by-Step Solution

Answer

\(u = \frac{\pi}{6} + 2k\pi\) and \(u = \frac{5\pi}{6} + 2k\pi\), where \(k\) is an integer.

Step 1: Understand the Zero Product Property

The given equation is \((2 \sin u - 1)(\cos u - \sqrt{2}) = 0\). According to the Zero Product Property, if the product of two factors is zero, then at least one of the factors must be zero. This gives us two separate equations to solve: \(2 \sin u - 1 = 0\) or \(\cos u - \sqrt{2} = 0\).

Step 2: Solve for \(u\) in the First Equation

First, solve \(2 \sin u - 1 = 0\) for \(u\). Add 1 to both sides to get \(2 \sin u = 1\). Then, divide both sides by 2 to find \(\sin u = \frac{1}{2}\). The solutions for \(u\) where \(\sin u = \frac{1}{2}\) are \(u = \frac{\pi}{6} + 2k\pi\) and \(u = \frac{5\pi}{6} + 2k\pi\), where \(k\) is any integer.

Step 3: Analyze the Second Equation

Now consider the second equation: \(\cos u - \sqrt{2} = 0\). This implies \(\cos u = \sqrt{2}\). However, the range of the cosine function is only between -1 and 1. Since \(\sqrt{2} \approx 1.41 > 1\), there is no real number \(u\) for which this equation holds. Therefore, this part of the equation provides no solutions.

Step 4: Combine Solutions

Combine the solutions from the first equation, as there were no solutions from the second. The solutions are \(u = \frac{\pi}{6} + 2k\pi\) and \(u = \frac{5\pi}{6} + 2k\pi\) for any integer \(k\).

Key Concepts

Zero Product Property

The Zero Product Property is crucial for solving equations where a product is set to zero. It states that if the product of two or more factors is zero, then at least one of the factors must be zero. For instance, if you have an equation like

• \((2 \, \sin \, u - 1)(\cos \, u - \sqrt{2}) = 0\),

you apply this property to break the equation into two separate equations:

• \(2 \, \sin \, u - 1 = 0\)
• \(\cos \, u - \sqrt{2} = 0\)

By treating each factor as potentially being zero, you can find solutions more easily. It is an effective way to reduce complex equations into simpler ones. Learning how to use this property effectively can greatly streamline the process of solving trigonometric and other types of polynomial equations.

Sine Function

The sine function, represented as \(\sin \, u\), is a fundamental trigonometric function. It describes the y-coordinate of a point on the unit circle corresponding to an angle \(u\). The sine function has the following properties:

• It is periodic with a period of \(2\pi\), which means the function repeats its values every \(2\pi\) radians.
• The range of \(\sin \, u\) is \([-1,1]\).
• Key values occur at specific angles: \(\sin \, \frac{\pi}{6} = \frac{1}{2}\) and \(\sin \, \frac{5\pi}{6} = \frac{1}{2}\).

When solving the equation \(2\,\sin \, u - 1 = 0\), you isolate sine to find \(\sin \, u = \frac{1}{2}\). This gives solutions at angles \(u = \frac{\pi}{6} + 2k\pi\) and \(u = \frac{5\pi}{6} + 2k\pi\) for any integer \(k\), due to the periodic nature of sine.

Cosine Function

The cosine function, expressed as \(\cos \, u\), is another essential trigonometric function. It gives the x-coordinate of a point on the unit circle corresponding to the angle \(u\). Important characteristics of the cosine function include:

• It has a period of \(2\pi\), repeating its values every \(2\pi\) radians, just like the sine function.
• The range is \([-1,1]\).
• Common values are at angles \(\cos \, 0 = 1\) and \(\cos \, \pi = -1\).

In the equation \(\cos \, u - \sqrt{2} = 0\), solving for \(u\) gives \(\cos \, u = \sqrt{2}\). However, this is impossible as \(\sqrt{2} \approx 1.41\), which is outside the range of \([-1,1]\). Therefore, there are no solutions for \(u\) from this part of the equation. Understanding the range and behavior of the cosine function is critical when evaluating the feasibility of solutions in trigonometric equations.