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Magazine Chapter 13: Problem 6
In \(3-14,\) find the exact values of \(\theta\) in the interval \(0^{\circ} \leq \theta < 360^{\circ}\) that satisfy each equation. $$ 2 \sin \theta+1=\csc \theta $$
Short Answer
Expert verified
The solutions are \(\theta = 30^\circ\), \(150^\circ\), and \(270^\circ\).
Step by step solution
01
Transform the Equation
The equation to solve is \(2 \sin \theta + 1 = \csc \theta\). Begin by expressing \(\csc \theta\) in terms of \(\sin \theta\):\[ \csc \theta = \frac{1}{\sin \theta} \]Substitute this identity into the equation:\[ 2 \sin \theta + 1 = \frac{1}{\sin \theta} \]
02
Create a Common Equation
Multiply both sides by \(\sin \theta\) to eliminate the fraction:\[ 2 \sin^2 \theta + \sin \theta = 1 \]
03
Rearrange the Equation
Reorganize the equation into a standard quadratic form:\[ 2 \sin^2 \theta + \sin \theta - 1 = 0 \]
04
Factor the Quadratic
Attempt to factor the quadratic equation:\[ (2 \sin \theta - 1)(\sin \theta + 1) = 0 \]This gives us potential solutions for \(\sin \theta\):1. \(2 \sin \theta - 1 = 0\) 2. \(\sin \theta + 1 = 0\)
05
Solve for Sine Values
Solve each of the factored equations for \(\sin \theta\):- From \(2 \sin \theta - 1 = 0\), we have: \[ \sin \theta = \frac{1}{2} \]- From \(\sin \theta + 1 = 0\), we have:\[ \sin \theta = -1 \]
06
Find Angles for \(\sin \theta = \frac{1}{2}\)
For \(\sin \theta = \frac{1}{2}\), the angles in the interval \(0^\circ \leq \theta < 360^\circ\) are \(\theta = 30^\circ\) and \(\theta = 150^\circ\).
07
Find Angle for \(\sin \theta = -1\)
For \(\sin \theta = -1\), the angle in the interval \(0^\circ \leq \theta < 360^\circ\) is \(\theta = 270^\circ\).
08
Conclusion
The angles that satisfy the original equation \(2 \sin \theta + 1 = \csc \theta\) within the interval are \(\theta = 30^\circ\), \(\theta = 150^\circ\), and \(\theta = 270^\circ\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sine Function
The sine function, denoted as \( \sin \theta \), is a fundamental trigonometric function derived from the opposite side's length over the hypotenuse in a right triangle. It plays a vital role in connecting angles to ratios of side lengths.
Key characteristics of the sine function:
Key characteristics of the sine function:
- *Range*: The sine function values range from -1 to 1.
- *Periodicity*: It has a periodic nature, repeating every 360掳 or \(2\pi\) radians.
- *Symmetry*: It is an odd function, which means \( \sin(-\theta) = -\sin(\theta) \).
Cosecant Function
The cosecant function, denoted as \( \csc \theta \), is the reciprocal of the sine function. This means \( \csc \theta = \frac{1}{\sin \theta} \). It represents the ratio of the hypotenuse to the opposite side in a right-angled triangle.
Key features of the cosecant function include:
Key features of the cosecant function include:
- *Range*: Since it is the reciprocal of sine, it does not exist where \( \sin \theta = 0 \). The range is \(( -\infty, -1] \cup [1, \infty )\).
- *Periodicity*: Like sine, the cosecant function is periodic with a period of 360掳 or \(2\pi\) radians.
- *Vertical asymptotes*: Occur at integer multiples of \( \pi \) where the sine function equals zero.
Quadratic Equation
A quadratic equation is any equation that can be arranged in the form \( ax^2 + bx + c = 0 \). In the context of trigonometric functions, it often arises when rearranging and simplifying expressions with squared trigonometrics, such as \( 2 \sin^2 \theta + \sin \theta - 1 = 0 \).
Solving a quadratic equation involves:
Solving a quadratic equation involves:
- *Factoring*: Expressing the quadratic as a product of two binomials. For \( (2 \sin \theta - 1)(\sin \theta + 1) = 0 \), we have simplified the equation down.
- *Quadratic formula*: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), is another method if factoring is not feasible.
Angle Solution
Finding the angle solutions to trigonometric equations involves determining the specific angles that satisfy the equation within a given interval, typically \(0^{\circ} \leq \theta < 360^{\circ}\).
To solve for angles:
To solve for angles:
- *Determine possible sine values*: From the solutions derived for \( \sin \theta \). In our example, these values were \( \sin \theta = \frac{1}{2} \) and \( \sin \theta = -1 \).
- *Reference the unit circle*: Identify angles where these sine values occur. \( \sin \theta = \frac{1}{2} \) corresponds to \(30^{\circ}\) and \(150^{\circ}\), while \( \sin \theta = -1 \) corresponds to \(270^{\circ}\).
