Problem 39
Question
Solve the given equation. $$\left(\tan ^{2} \theta-4\right)(2 \cos \theta+1)=0$$
Step-by-Step Solution
Verified Answer
The solutions are \(\theta = \arctan(2) + k\pi\), \(\theta = \arctan(-2) + k\pi\), \(\theta = \frac{2\pi}{3} + 2k\pi\), and \(\theta = \frac{4\pi}{3} + 2k\pi\) for any integer \(k\).
1Step 1: Analyze the Equation
The given equation is \((\tan^2 \theta - 4)(2 \cos \theta + 1) = 0\). This is a product of two factors set to zero, which means either \(\tan^2 \theta - 4 = 0\) or \(2 \cos \theta + 1 = 0\) must be true.
2Step 2: Solve \(\tan^2 \theta - 4 = 0\)
Start by solving the equation \(\tan^2 \theta - 4 = 0\). Move 4 to the other side: \(\tan^2 \theta = 4\). Taking the square root on both sides gives \(\tan \theta = \pm 2\). The solutions for \(\theta\) occur at: \(\theta = \arctan(2) + k\pi\) and \(\theta = \arctan(-2) + k\pi\), where \(k\) is any integer.
3Step 3: Solve \(2 \cos \theta + 1 = 0\)
Solve the second factor: \(2 \cos \theta + 1 = 0\). Subtract 1 from both sides: \(2 \cos \theta = -1\). Then divide by 2: \(\cos \theta = -\frac{1}{2}\). The solutions for \(\theta\) occur at: \(\theta = \frac{2\pi}{3} + 2k\pi\) and \(\theta = \frac{4\pi}{3} + 2k\pi\), where \(k\) is any integer.
4Step 4: Combine the Solutions
The solutions to the original equation are the union of the solutions from Steps 2 and 3. Therefore, \(\theta\) can be \(\arctan(2) + k\pi\), \(\arctan(-2) + k\pi\), \(\frac{2\pi}{3} + 2k\pi\), or \(\frac{4\pi}{3} + 2k\pi\) for any integer \(k\).
Key Concepts
Solving EquationsTrigonometric FunctionsTangent and Cosine Relations
Solving Equations
When dealing with equations like \[(\tan ^{2} \theta-4)(2 \cos \theta+1)=0\], it's important to recognize that the equation is in a "product equals zero" form. This means that for the entire product to be zero, at least one of the factors must be zero.
To solve such equations, we break them down into simpler parts:
To solve such equations, we break them down into simpler parts:
- Start by identifying the individual factors: \(\tan^2 \theta - 4\) and \(2 \cos \theta + 1\).
- Set each factor equal to zero separately to find possible solutions.
- For the first factor: \(\tan^2 \theta - 4 = 0\).
- For the second factor: \(2 \cos \theta + 1 = 0\).
Trigonometric Functions
Trigonometric functions are essential tools in mathematics, particularly when dealing with angles and triangles. In the context of this exercise, the trigonometric functions involved are the tangent and cosine functions.
The tangent function, \(\tan \theta\), relates the sine and cosine functions as \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). The square of this function, \(\tan^2 \theta = 4\), leads to solutions by taking square roots, which introduces the possibility of both positive and negative results.
Similarly, the cosine function, \(\cos \theta\), is a basic trigonometric function representing the ratio of the adjacent side to the hypotenuse in a right-angled triangle. Solving \(2 \cos \theta + 1 = 0\) requires manipulating the equation to express \(\cos \theta\) in terms of standard angles. Understanding these functions deeply helps in solving and interpreting the behavior of trigonometric equations.
The tangent function, \(\tan \theta\), relates the sine and cosine functions as \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). The square of this function, \(\tan^2 \theta = 4\), leads to solutions by taking square roots, which introduces the possibility of both positive and negative results.
Similarly, the cosine function, \(\cos \theta\), is a basic trigonometric function representing the ratio of the adjacent side to the hypotenuse in a right-angled triangle. Solving \(2 \cos \theta + 1 = 0\) requires manipulating the equation to express \(\cos \theta\) in terms of standard angles. Understanding these functions deeply helps in solving and interpreting the behavior of trigonometric equations.
Tangent and Cosine Relations
The relationship between tangent and cosine functions is pivotal for solving the given equation. In this specific exercise, we explore the behavior of both functions under specific conditions.
When \(\tan^2 \theta = 4\), we know that \(\tan \theta = \pm 2\), leading directly to \(\theta = \arctan(2) + k\pi\) or \(\theta = \arctan(-2) + k\pi\) for integer \(k\). This takes into account the tangent function's periodicity, repeating every \(\pi\) units.
On the other hand, solving \(\cos \theta = -\frac{1}{2}\) involves identifying angles where the cosine function yields \(-\frac{1}{2}\). This occurs at angles like \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\), though the function repeats every \(2\pi\). Both solutions reinforce the properties of tangent and cosine functions, illustrating how periodic and specific they are with respect to angle measurements.
When \(\tan^2 \theta = 4\), we know that \(\tan \theta = \pm 2\), leading directly to \(\theta = \arctan(2) + k\pi\) or \(\theta = \arctan(-2) + k\pi\) for integer \(k\). This takes into account the tangent function's periodicity, repeating every \(\pi\) units.
On the other hand, solving \(\cos \theta = -\frac{1}{2}\) involves identifying angles where the cosine function yields \(-\frac{1}{2}\). This occurs at angles like \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\), though the function repeats every \(2\pi\). Both solutions reinforce the properties of tangent and cosine functions, illustrating how periodic and specific they are with respect to angle measurements.
Other exercises in this chapter
Problem 39
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