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Ch. 6 - Inverse Circular Functions and Trigonometric Equations
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161Not the one you use?Change textbook
Chapter 7, Problem 17

Solve each equation for x, where x is restricted to the given interval.
y = ―4 + 2 sin x , for x in [―π/2. π/2]

Verified step by step guidance
1
Rewrite the equation to isolate the sine term: add 4 to both sides to get \(y + 4 = 2 \sin x\).
Divide both sides of the equation by 2 to solve for \(\sin x\): \(\sin x = \frac{y + 4}{2}\).
Recall that the sine function outputs values only between -1 and 1, so ensure that \(\frac{y + 4}{2}\) lies within this range for solutions to exist.
Use the inverse sine function to solve for \(x\): \(x = \arcsin\left(\frac{y + 4}{2}\right)\).
Since \(x\) is restricted to the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\), the principal value of \(\arcsin\) will give the solution(s) within this interval.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Sine Function and Its Properties

The sine function, sin(x), is a periodic trigonometric function that oscillates between -1 and 1. Understanding its behavior within a specific interval, such as [-π/2, π/2], is crucial because it is monotonic and covers all values from -1 to 1 in this range, simplifying equation solving.
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Solving Trigonometric Equations

Solving trigonometric equations involves isolating the trigonometric function and finding all angle values that satisfy the equation within the given domain. This often requires using inverse trigonometric functions and considering the function's periodicity and restrictions on the interval.
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Domain Restrictions and Interval Considerations

Restricting the variable x to a specific interval, such as [-π/2, π/2], limits the possible solutions to those within that range. This is important because trigonometric functions are periodic, and multiple solutions may exist outside the interval, but only those within the domain are valid.
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