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

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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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Concetti chiave

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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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4:03

Inverse Sine

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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Domain and Range of Function Transformations

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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