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

검증된 단계별 안내

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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주요 개념

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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.

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.

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.