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Ch. 3 - Trigonometric Identities and Equations
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 3.5.45

Exercises 39–52 involve trigonometric equations quadratic in form. Solve each equation on the interval [0, 2𝝅). sin² θ - 1 = 0

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Recognize that the equation is quadratic in form with respect to \(\sin \theta\). The given equation is \(\sin^{2} \theta - 1 = 0\).
Rewrite the equation to isolate the squared term: \(\sin^{2} \theta = 1\).
Take the square root of both sides, remembering to consider both positive and negative roots: \(\sin \theta = \pm 1\).
Determine the values of \(\theta\) in the interval \([0, 2\pi)\) where \(\sin \theta = 1\) or \(\sin \theta = -1\). Recall the unit circle values for sine.
List all solutions for \(\theta\) in \([0, 2\pi)\) that satisfy the equation based on the sine values found.

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Quadratic Form in Trigonometric Equations

A trigonometric equation is quadratic in form when it can be expressed similarly to a quadratic equation, such as involving terms like sin²θ or cos²θ. Recognizing this allows the use of algebraic methods like substitution to solve the equation efficiently.
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Solving Basic Trigonometric Equations

Solving equations like sin²θ - 1 = 0 involves isolating the trigonometric function and finding all angles θ within the given interval that satisfy the equation. This requires understanding the values of sine and cosine on the unit circle.
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Interval Restriction and Solution Sets

When solving trigonometric equations, solutions are often restricted to a specific interval, such as [0, 2π). It is important to find all solutions within this range, considering the periodic nature of trigonometric functions.
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