Problem 86 - Precalculus Essentials

Question

Use a calculator to solve each equation, correct to four decimal places, on the interval \([0,2 \pi)\) $$\sin x=0.7392$$

Step-by-Step Solution

Answer

The solutions to the equation \( \sin x =0.7392 \) on the interval \([0,2 \pi)\) are approximately \(x = \arcsin(0.7392)\) and \(x = \pi + \arcsin(0.7392)\), rounded to four decimal places.

Step 1: Apply the inverse sine function

The first step is to apply the inverse sine function, or \(\arcsin\), to the value of 0.7392. This is done using a calculator.

Step 2: Find the second solution in interval

Because the sine function repeats every \(2\pi\) radians, there is another solution within the interval \([0,2 \pi)\). This can be found by adding \(\pi\) to the initial solution obtained in step 1.

Step 3: Round results

Finally, round both solutions obtained in steps 1 and 2 to four decimal places, as the problem statement requests.

Key Concepts

Inverse Functions

Inverse functions play a crucial role in solving trigonometric equations by reversing the processes of the direct functions. For trigonometric equations involving the sine function, the inverse function used is the arcsine, denoted as \( \arcsin \). This function helps to determine the angle whose sine is a given number. When you see an equation like \( \sin x = 0.7392 \), you can find the angle \( x \) by applying the inverse sine function. You would input \( \arcsin(0.7392) \) into a calculator to receive a primary angle measurement. It's important to remember that calculators are typically set to provide results in radians or degrees, so make sure your output mode matches your expected answer format. This method helps isolate \( x \) as it's the angle measurement that, when passed through the sine function, would yield the original equation's result.

Intervals

An interval specifies a range of values. When solving trigonometric equations, the specified interval defines where solutions should be searched within. In this exercise, the interval given is \([0, 2\pi)\).The interval \([0, 2\pi)\) means all real numbers starting from 0 up to, but not including, \( 2\pi \). When working in radians, \( 2\pi \) represents a full circle, so this interval dictates that all solutions should be found within one complete rotation of the unit circle.This constraint is important because trigonometric functions, like sine, are periodic. It means they repeat their values in regular intervals. In this case of sine, the function repeats every \( 2\pi \) radians. That repetition allows multiple answers for \( x \) within any two consecutive intervals of \( 2\pi \). Keeping the correct interval helps ensure that all possible solutions within the first cycle are considered.

Sine Function

The sine function is one of the fundamental trigonometric functions, traditionally defined in the context of a right-angled triangle. It relates an angle to the ratio of the length of the opposite side to the hypotenuse. For any given angle \( x \), \( \sin x \) provides the corresponding sine value.For solving equations like the one given—\( \sin x = 0.7392 \)—understanding its properties is crucial. The sine function is known for its smooth wave-like pattern, which cycles every \( 2\pi \) radians, similar to a continuous oscillation.Some key points to remember about the sine function:

• The range of \( \sin x \) is from -1 to 1.
• It is periodic, with a period of \( 2\pi \) radians.
• The values repeat, meaning an equation like this will often have more than one valid solution within any given interval.

Understanding these characteristics helps solve trigonometric equations efficiently, especially when determining additional solutions by considering the cycle of the sine function.