Textbook Question
Solve each equation for x, where x is restricted to the given interval.
y = ― 2 cos 5x , for x in [0, π/5]
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5. Inverse Trigonometric Functions and Basic Trigonometric Equations
Inverse Sine, Cosine, & Tangent
Problem 15
Textbook Question
Solve each equation for x, where x is restricted to the given interval.
y = sin x ―2 , for x in [―π/2. π/2]
Verified step by step guidance1
Identify the given equation: \(y = \sin x - 2\), and the interval for \(x\) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Rewrite the equation to isolate \(\sin x\): set \(y = 0\) to find the values of \(x\) where \(\sin x - 2 = 0\), which gives \(\sin x = 2\).
Recall the range of the sine function: \(\sin x\) can only take values between \(-1\) and \(1\) for all real \(x\).
Since \(2\) is outside the range \([-1, 1]\), conclude that there are no real solutions for \(x\) in the interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) where \(\sin x = 2\).
Therefore, the equation \(y = \sin x - 2\) has no solutions for \(x\) in the given interval.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Inverse Sine Function (Arcsin)
The inverse sine function, denoted as arcsin or sin⁻¹, is used to find the angle x when given the sine value y. It returns values within the principal range of [−π/2, π/2], which matches the interval restriction in the problem.
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Domain and Range of the Sine Function
The sine function outputs values between -1 and 1 for all real inputs. Understanding this range is crucial because the equation y = sin x - 2 shifts the sine values down by 2, affecting the possible values of y and the solvability of the equation.
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Interval Restrictions on the Variable
The problem restricts x to the interval [−π/2, π/2], which is the principal domain for the arcsin function. This restriction ensures a unique solution for x when solving the equation and aligns with the range of the inverse sine.
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