Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 6: Problem 50
Find the solutions of the equation that are in the interval \([\mathbf{0}, \mathbf{2} \pi)\). $$2-\cos ^{2} x=4 \sin ^{2} \frac{1}{2} x$$
Short Answer
Expert verified
The solutions are \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).
Step by step solution
01
Use a Trigonometric Identity
Recall the trigonometric identity \( \cos^2{x} = 1 - 2\sin^2{\left(\frac{x}{2}\right)} \). We will use this identity to substitute \(\cos^2{x}\) in the given equation. So, we have \( 2 - (1 - 2\sin^2{\left(\frac{x}{2}\right)} = 4\sin^2{\left(\frac{1}{2}x\right)} \).
02
Simplify the Equation
Simplify the equation obtained in Step 1: \[ 2 - (1 - 2\sin^2{\left(\frac{x}{2}\right)}) = 4\sin^2{\left(\frac{1}{2}x\right)} \]This becomes \[ 1 + 2\sin^2{\left(\frac{x}{2}\right)} = 4\sin^2{\left(\frac{1}{2}x\right)} \]
03
Simplify Further
Reorganize the equation: \[ 1 = 4\sin^2{\left(\frac{1}{2}x\right)} - 2\sin^2{\left(\frac{x}{2}\right)} \] This simplifies to:\[ 1 = 2\sin^2{\left(\frac{x}{2}\right)} \]Now divide both sides by 2 to isolate the sine term:\[ \sin^2{\left(\frac{x}{2}\right)} = \frac{1}{2} \]
04
Solve for \( \sin{\left(\frac{x}{2}\right)} \)
Take the square root of both sides:\[ \sin{\left(\frac{x}{2}\right)} = \pm\frac{1}{\sqrt{2}} \]Since \( \sin{\left(\frac{x}{2}\right)} = \pm\frac{\sqrt{2}}{2} \), \( \frac{x}{2} \) could be angles \( \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \) or \( \frac{7\pi}{4} \).
05
Solve for \( x \)
Multiply each angle by 2 to find the solutions for \( x \):\[ x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \text{and } \frac{7\pi}{2} \]Notice that some solutions \( \frac{5\pi}{2} \) and \( \frac{7\pi}{2} \) fall outside \([0, 2\pi) \).Adjust them by subtracting \( 2\pi \) if necessary:\[ \frac{5\pi}{2} - 2\pi = \frac{\pi}{2}, \quad \frac{7\pi}{2} - 2\pi = \frac{3\pi}{2} \]
06
Final Solutions
The valid solutions in the interval \([0, 2\pi) \) are \( x = \frac{\pi}{2}, \text{and } \frac{3\pi}{2} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Identities
In trigonometry, identities are equations that hold true for all valid values of the variables involved. These identities are particularly useful when solving trigonometric equations, as they can transform complex expressions into simpler ones. In this exercise, we make use of the trigonometric identity \[\cos^2{x} = 1 - 2\sin^2{\left(\frac{x}{2}\right)}\] This identity comes from the half-angle formulas, which relate the cosine and sine functions of an angle to the trigonometric functions of half that angle. By substituting this identity into the original equation, we reduce the complexity of the equation significantly. This step is essential as it converts all terms to a single function, which is easier to solve.
Interval Solutions
When solving trigonometric equations, it is crucial to find solutions within a specified interval. For this problem, the interval \([0, 2\pi)\) is specified, meaning we seek solutions that fall within one complete cycle of the trigonometric functions. After finding the general solutions for the variable, we must check that they lie within this interval. This often requires adjusting angles by adding or subtracting multiples of \(2\pi\) until they fit the interval constraints. In this exercise, solutions such as\(\frac{5\pi}{2}\) and \(\frac{7\pi}{2}\) were adjusted by subtracting \(2\pi\) to fit within \([0, 2\pi)\). This step ensures that only the applicable and valid solutions are considered.
Sine and Cosine Functions
Understanding the sine and cosine functions is foundational when dealing with trigonometric equations. The sine function \(\sin{x}\) and cosine function \(\cos{x}\) are periodic, meaning they repeat values in regular intervals. Their periodic nature is why trigonometric equations often have multiple solutions within given intervals. Both functions have ranges between \([-1, 1]\), and their behavior within any interval can be predicted using various identities and properties. In this exercise, we centered our solutions around the sine function's half-angle property, converting the equation into a form where all terms are expressed through this function: \[\sin^2\left(\frac{x}{2}\right)\]. By applying the property of periodic repetition, we gather potential solutions which then need validation against interval restrictions. This approach allows for a systematic and logical path to finding the correct solutions to trigonometric equations.
