Problem 21 - Precalculus with Limits

Question

In Exercises 11-24, solve the equation. \( 2 \sin^2 2x = 1 \)

Step-by-Step Solution

Answer

The solution to the given equation is \( x = \pm \frac{\pi}{8} + n\pi, \pm \frac{5\pi}{8} + n\pi \) where \( n \) is an integer.

Step 1: Simplify the equation

The first step is to simplify the equation by dividing both sides by 2. This results in \( \sin^2 2x = \frac{1}{2} \).

Step 2: Derive the sin function

Since we now have \(\sin^2 2x = \frac{1}{2}\), which means \( \sin 2x = \pm \sqrt{\frac{1}{2}} \). Note that \(\sqrt{1/2}\) can be simplified to \(\frac{1}{\sqrt{2}}\) or \(\frac{\sqrt{2}}{2}\) using the rule that the square root of a fraction is equal to the square root of the numerator divided by the square root of the denominator.

Step 3: Apply the trigonometric identity

The fact that \(\sin 2x = \pm \frac{\sqrt{2}}{2}\) tells us that 2x is equal to specific angles where the sine value is \(\pm \frac{\sqrt{2}}{2}\), which is \(\pm \frac{\pi}{4}\), \(\pm \frac{5\pi}{4}\), \( \pm \frac{9\pi}{4} \) and \( \pm \frac{13\pi}{4} \), etc. So, we have \( 2x = \pm \frac{\pi}{4} + 2n\pi, \pm \frac{5\pi}{4} + 2n\pi \) where \( n \) is an integer since trigonometric functions are periodic. Then, x has the general solution \( x = \pm \frac{\pi}{8} + n\pi, \pm \frac{5\pi}{8} + n\pi \) by dividing the whole equation by 2.

Key Concepts

trigonometric identities

Trigonometric identities are mathematical equations that depict relationships between the trigonometric functions such as sine, cosine, and tangent. These identities are crucial for simplifying expressions and solving equations involving trigonometric functions.

One common identity is the Pythagorean identity, which tells us that \( \sin^2\theta + \cos^2\theta = 1 \). This is very useful when dealing with squares of sine or cosine functions because it gives us a way to express one in terms of the other. Another useful identity is the double-angle identity, notably \( \sin 2x = 2 \sin x \cos x \), which helps in transforming the equation into forms that are more manageable. In the exercise, we simplified the equation by using square roots. Recognizing identities like these can help you solve problems where these trigonometric functions are involved.

Make sure to memorize key trigonometric identities and understand their applications. This will improve your ability to manipulate and solve equations mathematically.

sine function

The sine function is one of the primary trigonometric functions and is fundamental in trigonometry. It is often represented as \( \sin(\theta) \), where \( \theta \) is an angle in radians or degrees.

The range of the sine function is between -1 and 1, making it very predictable and manageable to work with. In the exercise, when we resolved \( \sin^2 2x = \frac{1}{2} \), we essentially needed to find which angles, when plugged into the sine function, resulted in values of \( \pm \frac{\sqrt{2}}{2} \).

• For \( \sin(\theta) = \frac{\sqrt{2}}{2} \), typical angles include \( \frac{\pi}{4} \) and \( \frac{5\pi}{4} \).
• For \( \sin(\theta) = -\frac{\sqrt{2}}{2} \), angles such as \( -\frac{\pi}{4} \) and \( -\frac{5\pi}{4} \) apply.

Understanding how the sine function behaves with respect to different angles allows you to easily solve trigonometric equations by converting the function value to its corresponding angles.

periodic functions

Trigonometric functions, like sine and cosine, are periodic, meaning they repeat their values in regular intervals. This property is vital when solving trigonometric equations as it helps in finding all possible solutions.

The sine function has a period of \(2\pi\); this means that every \(2\pi\) radians, the output values begin to repeat. In the original exercise, to find all solutions for \( x \), it was necessary to account for the periodic nature of the sine function:

• Notice that for \( 2x \), values like \( \frac{\pi}{4} \), \( \frac{5\pi}{4} \), etc., come from the sine value of \( \pm \frac{\sqrt{2}}{2} \).
• Adding multiples of the period, \( 2\pi \), modifies these angles to find all solutions (e.g., \( 2x = \frac{\pi}{4} + 2n\pi \)).
When you solve for \( x \), consider these periodic properties, dividing by two to adjust for \( 2x \), resulting in \( x = \frac{\pi}{8} + n\pi \). By keeping periodicity in mind, you ensure that every valid solution over the range of the function is found.