Problem 6 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation over the interval \([0,2 \pi)\) $$\sin x \cos x=\frac{1}{4}$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \) and \( \frac{17\pi}{12} \).

Step 1: Use Trigonometric Identity

Start by using the trigonometric identity \( 2\sin x \cos x = \sin 2x \). So, the given equation becomes \( \frac{1}{2} \sin 2x = \frac{1}{4} \). Multiply both sides by 2 to simplify the equation to \( \sin 2x = \frac{1}{2} \).

Step 2: Solve for Angle

Determine the angles \( 2x \) that satisfy the equation \( \sin 2x = \frac{1}{2} \). The angles for which the sine is \( \frac{1}{2} \) within one full cycle are \( 2x = \frac{\pi}{6} \) and \( 2x = \frac{5\pi}{6} \), since sine is positive in the first and second quadrants.

Step 3: Solve for x

Solve for \( x \) in the equations \( 2x = \frac{\pi}{6} + 2k\pi \) and \( 2x = \frac{5\pi}{6} + 2k\pi \), where \( k \) is an integer. This gives \( x = \frac{\pi}{12} \) and \( x = \frac{5\pi}{12} \) when \( k = 0 \).

Step 4: Additional Cycle Consideration

We also consider if there are any additional solutions within the interval \([0, 2\pi)\) by substituting \( k = 1 \), which yields additional values \( x = \frac{13\pi}{12} \) and \( x = \frac{17\pi}{12} \). Verify each is within the interval.

Step 5: Final Solution Set

Thus, the solutions in the interval \([0, 2\pi)\) are \( x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \) and \( \frac{17\pi}{12} \).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools that help simplify expressions and solve equations involving trigonometric functions. They are like little shortcuts and formulas connecting different trigonometric functions to one another. One important identity is the product-to-sum identity, such as:

• \( 2 \sin x \cos x = \sin 2x \)

This particular identity helps us condense the product of \( \sin x \cos x \) into a single trigonometric function, \( \sin 2x \). By applying this identity, the equation becomes simpler and easier to solve.
Understanding these identities is crucial for simplifying complex problems, as it can reduce the number of unknowns and result in easier manipulation of the expressions.

Sine Function

The sine function \( \sin \) is one of the primary trigonometric functions and describes the vertical component of a point on the unit circle. With an amplitude between -1 and 1, it repeats its pattern every \( 2\pi \) radians (360 degrees) across the unit circle.

• It is positive in the first and second quadrants of the circle, where the angle ranges are \( 0 \leq x \leq \pi \).
• Key reference angles, such as \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \), help find where specific sinusoidal values occur.

These angles are essential when solving equations like \( \sin 2x = \frac{1}{2} \), as they allow us to identify \( 2x \) values by recognizing where these specific \( \sin \) values occur on the unit circle.

Angle Resolution

Resolving angles is about finding the precise angle values that satisfy a trigonometric equation. When given an equation like \( \sin 2x = \frac{1}{2} \), you first determine for which angles \( \sin \) reaches \( \frac{1}{2} \).
Within one cycle of the sine function, these angles are:

• \( 2x = \frac{\pi}{6} \)
• \( 2x = \frac{5\pi}{6} \)

To resolve these angles, you solve each equation for \( x \) by dividing by 2, considering angles in radians are often within the interval of \( [0, 2\pi) \).
Solving for \( x \) involves unwinding the transformation from \( 2x \) to \( x \), ensuring all relevant cycles are accounted for within a given interval.

Interval Solutions

In trigonometry, solving an equation over a specific interval means finding all the possible solutions that fit within a set range of values. For this particular problem, we seek solutions of \( x \) in the interval \([0, 2\pi)\).

• First, find the base solutions: \( x = \frac{\pi}{12} \) and \( x = \frac{5\pi}{12} \).
• Consider additional cycles: When \( k = 1 \), \( 2\pi \) is added to the angle but divided by 2 during conversions to \( x \). This reveals more solutions such as \( x = \frac{13\pi}{12} \) and \( x = \frac{17\pi}{12} \).

All these solutions need to be checked to ensure they fall within \( [0, 2\pi) \). This approach ensures the final solution set is complete and encompasses all angles valid in the stipulated range.