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Ch. 6 - Inverse Circular Functions and Trigonometric Equations
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161, 9780135440780, 9780136881117, 9780135440827, 9780135924891Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 6.3.19

Solve each equation in x over the interval [0, 2π) and each equation in θ over the interval [0°, 360°). Give exact solutions.


sin 3θ = -1

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1
Recognize that the equation is \( \sin 3\theta = -1 \). We want to find all values of \( \theta \) in the interval \( [0^\circ, 360^\circ) \) such that this holds true.
Recall that \( \sin x = -1 \) at specific angles. The sine function equals \( -1 \) at \( x = 270^\circ + 360^\circ k \), where \( k \) is any integer.
Set the inside of the sine function equal to these angles: \( 3\theta = 270^\circ + 360^\circ k \). This accounts for all possible solutions for \( 3\theta \).
Solve for \( \theta \) by dividing both sides by 3: \( \theta = \frac{270^\circ + 360^\circ k}{3} = 90^\circ + 120^\circ k \).
Find all values of \( \theta \) within the interval \( [0^\circ, 360^\circ) \) by substituting integer values of \( k \) such that \( \theta \) remains in this range.

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Solving Trigonometric Equations

Solving trigonometric equations involves finding all angle values within a specified interval that satisfy the given equation. This requires understanding the periodic nature of trig functions and applying inverse functions to isolate the variable.
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How to Solve Linear Trigonometric Equations

Properties of the Sine Function

The sine function oscillates between -1 and 1 with a period of 2π radians (360°). Knowing where sine equals -1, specifically at 3π/2 radians (270°), helps identify principal solutions before considering periodic repetitions.
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Graph of Sine and Cosine Function

Interval and Multiple-Angle Considerations

When the argument of sine is multiplied by a factor (e.g., 3θ), the function's period changes, affecting the number of solutions within the original interval. Adjusting the interval accordingly and finding all solutions requires dividing the interval by the multiplier.
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Coterminal Angles
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