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Chapter 8: Problem 83

Solve each equation on the interval \(0 \leq \theta<2 \pi\) \(\sec ^{2} \theta+\tan \theta=0\)

Short Answer

Expert verified
There are no real solutions to the equation \sec^2 \theta + \tan \theta = 0 \in the interval \0 \text{鈮 \theta < 2\text{蟺}\.

Step by step solution

01

Rewrite using trigonometric identities

Recall that \(\text{{sec}} ^{2} \theta = 1 + \tan ^{2} \theta\). Substitute this into the given equation: \(1 + \tan ^{2} \theta + \tan \theta = 0. \) This simplifies to \(1 + \tan ^{2} \theta + \tan \theta = 0.\)
02

Form a quadratic equation

Let \(x = \tan \theta\). Rewrite the equation as a quadratic in terms of \(x\): \1 + x^2 + x = 0.\ This can be rewritten as \x^2 + x + 1 = 0.\
03

Solve the quadratic equation

To solve \(x^2 + x + 1 = 0\), use the quadratic formula \(x = \frac{{-b \text{卤} \text{鈭殅(b^2-4ac)}}{{2a}} \). Here, \(a = 1, b = 1, c = 1\). Substitute these values into the quadratic formula: \x = \frac{{-1 \text{卤} \text{鈭殅(1^2 - 4(1)(1))}}{{2(1)}} = \frac{{-1 \text{卤} \text{鈭殅(-3)}}{{2}}\. Since the discriminant \(1 - 4 = -3\) is negative, there are no real solutions for \(x\).
04

Conclude there are no real solutions

Since \(x = \tan \theta\) leads to a quadratic with a negative discriminant, \x = \tan \theta\ has no real solutions in \(0 \text{鈮 \theta < 2\text{蟺}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

trigonometric identities
Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. They simplify and connect trigonometric functions. In this exercise, we use the identity: \(\sec^2\theta = 1 + \tan^2\theta\). This relationship allows us to transform the original equation into a more manageable form. \(\sec\theta\) is the reciprocal of \(\cos\theta\), and understanding these relationships helps in rewriting and simplifying trigonometric equations. Identifying the right identities and knowing how to apply them is crucial in solving these problems effectively.
quadratic equations
A quadratic equation is any polynomial equation of the second degree, commonly represented as \(ax^2 + bx + c = 0\). In this exercise, we converted the trigonometric equation using the substitution \(x = \tan\theta\) to get \(x^2 + x + 1 = 0\). Solving quadratic equations generally involves factoring, completing the square, or using the quadratic formula. Each method has its advantages, but the quadratic formula is particularly useful when the equation does not factorize easily. Here, \(x = \frac{{-b \text{卤} \sqrt{b^2-4ac}}}{{2a}}\) gives us a direct way to find the solutions.
interval notation
Interval notation is a way of writing the set of all numbers between two endpoints. It includes the endpoints in the interval if they are specified with brackets. The given interval \(0 \leq \theta < 2\pi\) tells us to find solutions only within this range, which includes 0 but excludes \(2\pi\). This constraint is crucial as it restricts our solution set to one full rotation around the unit circle, ensuring that only valid angles within this span are considered.
discriminant
The discriminant in a quadratic equation \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\). It tells us the nature of the roots of the quadratic equation:
  • If \(b^2 - 4ac > 0\), the equation has two distinct real roots.
  • If \(b^2 - 4ac = 0\), it has exactly one real root (a repeated root).
  • If \(b^2 - 4ac < 0\), it has no real roots, only complex roots.
In our exercise, we find \(b^2 - 4ac = -3\), indicating no real solutions. Understanding the discriminant helps in quickly assessing whether the quadratic equation will have real solutions or if further steps involving complex numbers are required.

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Most popular questions from this chapter

Establish each identity. $$ \frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\tan \alpha+\tan \beta $$

Find the exact value of each expression. $$ \cos \left(\tan ^{-1} \frac{4}{3}+\cos ^{-1} \frac{5}{13}\right) $$

Find the exact value of each expression. $$ \cos \left[\tan ^{-1} \frac{5}{12}-\sin ^{-1}\left(-\frac{3}{5}\right)\right] $$

Convert \(6^{x}=y\) to an equivalent statement involving a logarithm.

Solve each equation on the interval \(0 \leq \theta<2 \pi\). $$ \cot \theta+\csc \theta=-\sqrt{3} $$
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