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Magazine Chapter 7: Problem 19
Find all solutions on the interval \([0,2 \pi)\) \(\sin ^{2} x=\frac{1}{4}\)
Short Answer
Expert verified
The solutions are \( x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \).
Step by step solution
01
Understand the Equation
The exercise requires us to solve the equation \( \sin^2 x = \frac{1}{4} \) within the interval \([0, 2\pi)\). The goal is to find all possible values of \(x\) within this interval that satisfy the equation.
02
Solve for \( \sin x \)
To solve for \( \sin x \), take the square root of both sides of the equation: \( \sin x = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2} \). This means we will solve for two cases: \( \sin x = \frac{1}{2} \) and \( \sin x = -\frac{1}{2} \).
03
Find Solutions for \( \sin x = \frac{1}{2} \)
\( \sin x = \frac{1}{2} \) at two angles within \([0, 2\pi)\): \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \). These are the solutions in the first and second quadrants where the sine function is positive.
04
Find Solutions for \( \sin x = -\frac{1}{2} \)
\( \sin x = -\frac{1}{2} \) at two angles within \([0, 2\pi)\): \( x = \frac{7\pi}{6} \) and \( x = \frac{11\pi}{6} \). These correspond to the third and fourth quadrants where the sine function is negative.
05
List All Solutions
The solutions to the equation \( \sin^2 x = \frac{1}{4} \) on the interval \([0, 2\pi)\) are \( x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \). These are the angles within the specified interval where the equation holds true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Unit Circle
The Unit Circle is a fundamental tool in trigonometry often used to solve trigonometric equations like the one in this exercise: \[ \sin^2 x = \frac{1}{4} \]The Unit Circle is a circle with a radius of one, centered at the origin
- We use it to visualize and find the cosine and sine of angles.
- The circle is divided into four quadrants, each with its own characteristics concerning the sine and cosine values.
- In the first quadrant, both sine and cosine values are positive. As the angle moves to the second quadrant, sine remains positive, but cosine becomes negative.
- Third-quadrant angles have both sine and cosine negative, whereas fourth-quadrant angles have negative sine and positive cosine values.
Sine Function
The Sine Function is one of the primary trigonometric functions and plays a crucial role in solving the equation provided:\[ \sin^2 x = \frac{1}{4} \]The sine of an angle measures the y-coordinate of a point on the Unit Circle. Here鈥檚 a breakdown of how it works:
- For each angle, the sine value corresponds to its vertical position above or below the x-axis.
- The function is periodic with a period of \( 2\pi \), meaning it repeats every \( 2\pi \).
- The sine function has specific values at certain angles: for example, \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \), and \( \sin \left( \frac{5\pi}{6} \right) = \frac{1}{2} \).
Interval Notation
Interval Notation is a mathematical tool used to specify a range of values that a variable can take. In this exercise, the interval\([0, 2\pi)\) dictates the set of potential solutions:
- The bracket "[" at the start means that the value 0 is included in the solutions (closed at 0).
- The parenthesis ")" at the end implies that \( 2\pi \) is not included (open at \( 2\pi \)).
- In this exercise, only angles from \( 0 \) to just before \( 2\pi \) are considered.
- Identifying the correct solutions applicable within this specified interval is key to solving the equation accurately.
