Problem 28
Question
Solve each equation for \(0 \leq \theta<2 \pi\). $$ \tan ^{2} \theta+\tan \theta=0 $$
Step-by-Step Solution
Verified Answer
The solutions for the equation are \( \theta = 0 , \pi , 3\pi/4 , 7\pi/4 \)
1Step 1: Rewrite the equation
Rewriting the equation \(\tan^{2}\theta + \tan\theta = 0\) in terms of \(\tan\theta\), gives: \(\tan\theta(\tan\theta + 1) = 0\)
2Step 2: Solve for theta using factoring
Having factored the equation in terms of \(\tan\theta\), next solve for \(\theta\) knowing that for an expression a*b=0, either a=0 or b=0. So either: \[\tan\theta = 0\: \text{or } \: \tan\theta + 1 = 0\] which gives: \(\tan\theta = 0\: \text{or } \: \tan\theta = -1\)
3Step 3: Find theta
Recalling that \(\tan\theta = \sin\theta / \cos\theta\) and knowing the periodic properties of the tangent and the interval \(0 \leq \theta < 2 \pi\), we find that: \(\theta = 0, \pi\) for \(\tan\theta = 0\), and \(\theta = \frac{3\pi}{4}, \frac{7\pi}{4}\) for \(\tan\theta = -1\)
4Step 4: Validation
Make sure that these answers are right by substituting them back into the equation. The above solutions hold and hence, they are valid values of \(\theta\)
Key Concepts
Understanding the Tangent FunctionStrategies for Solving Trigonometric EquationsDetermining Angle Solutions Within an Interval
Understanding the Tangent Function
The tangent function, denoted as \( \tan \theta \), is one of the fundamental functions in trigonometry. Unlike the sine and cosine functions, which are based on the ratios of the sides of a right triangle, the tangent is the ratio of the opposite side to the adjacent side. It can also be expressed in terms of sine and cosine as: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
A straightforward feature about the tangent function is its periodicity. This function is periodic with a period of \( \pi \), meaning that the function repeats its values every \( \pi \) radians. This property is important when solving trigonometric equations, as you'll frequently encounter multiple solutions within a given interval.
Another critical aspect of the tangent function is its vertical asymptotes, which occur at the angles where \( \cos \theta = 0 \). These asymptotes indicate that the function's value approaches infinity as it nears specific angles like \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \). In solving equations, awareness of these asymptotes helps avoid confusion about where tangent is undefined.
A straightforward feature about the tangent function is its periodicity. This function is periodic with a period of \( \pi \), meaning that the function repeats its values every \( \pi \) radians. This property is important when solving trigonometric equations, as you'll frequently encounter multiple solutions within a given interval.
Another critical aspect of the tangent function is its vertical asymptotes, which occur at the angles where \( \cos \theta = 0 \). These asymptotes indicate that the function's value approaches infinity as it nears specific angles like \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \). In solving equations, awareness of these asymptotes helps avoid confusion about where tangent is undefined.
Strategies for Solving Trigonometric Equations
Solving trigonometric equations often requires a mix of algebraic manipulation and an understanding of trigonometric identities. For the equation \( \tan^2 \theta + \tan \theta = 0 \), the first step is to factor it. Factoring is useful because it breaks down the problem into simpler parts that can each be solved individually. With this equation, you can rewrite it as:
\[ \tan \theta (\tan \theta + 1) = 0 \]
Once you have this factorization, apply the zero-product property. This property tells us that if a product of two expressions is zero, then at least one of the expressions must be zero. Therefore, we have two possible equations to solve:
- \( \tan \theta = 0 \)- \( \tan \theta = -1 \)
The next step involves finding all possible solutions for each case within the interval \( 0 \leq \theta < 2\pi \). Keep in mind the periodic nature of the tangent function. This means each equation might yield multiple solutions within the given interval.
\[ \tan \theta (\tan \theta + 1) = 0 \]
Once you have this factorization, apply the zero-product property. This property tells us that if a product of two expressions is zero, then at least one of the expressions must be zero. Therefore, we have two possible equations to solve:
- \( \tan \theta = 0 \)- \( \tan \theta = -1 \)
The next step involves finding all possible solutions for each case within the interval \( 0 \leq \theta < 2\pi \). Keep in mind the periodic nature of the tangent function. This means each equation might yield multiple solutions within the given interval.
Determining Angle Solutions Within an Interval
To find the angle solutions for \( \tan \theta = 0 \) and \( \tan \theta = -1 \) within \( 0 \leq \theta < 2\pi \), we must consider the specific angles where these conditions occur.
For \( \tan \theta = 0 \), recall that tangent equals zero when the sine of the angle is zero while the cosine is non-zero. This happens at angles \( \theta = 0 \) and \( \theta = \pi \).
For \( \tan \theta = -1 \), we look for angles where both sine and cosine have equal magnitude but opposite signs. These occur at \( \theta = \frac{3\pi}{4} \) and \( \theta = \frac{7\pi}{4} \).
When working through these solutions, it's crucial to validate each by substituting back into the original equation. This ensures that no computational errors have been made and confirms that all solutions are valid for the initial problem statement.
For \( \tan \theta = 0 \), recall that tangent equals zero when the sine of the angle is zero while the cosine is non-zero. This happens at angles \( \theta = 0 \) and \( \theta = \pi \).
For \( \tan \theta = -1 \), we look for angles where both sine and cosine have equal magnitude but opposite signs. These occur at \( \theta = \frac{3\pi}{4} \) and \( \theta = \frac{7\pi}{4} \).
When working through these solutions, it's crucial to validate each by substituting back into the original equation. This ensures that no computational errors have been made and confirms that all solutions are valid for the initial problem statement.
Other exercises in this chapter
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