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Ch. 6 - Inverse Circular Functions and Trigonometric Equations
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161Not the one you use?Change textbook
All textbooksLial 12th EditionCh. 6 - Inverse Circular Functions and Trigonometric EquationsProblem 6.RE.49
Chapter 7, Problem 6.RE.49

Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth of a degree, as appropriate.
3 cos² θ + 2 cos θ - 1 = 0

Verified step by step guidance
1
Recognize that the given equation is a quadratic in terms of \( \cos \theta \): \( 3 \cos^{2} \theta + 2 \cos \theta - 1 = 0 \). To solve it, let \( x = \cos \theta \), so the equation becomes \( 3x^{2} + 2x - 1 = 0 \).
Use the quadratic formula to solve for \( x \): \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a = 3 \), \( b = 2 \), and \( c = -1 \).
Calculate the discriminant \( \Delta = b^{2} - 4ac \) and then find the two possible values for \( x = \cos \theta \).
For each value of \( x \), determine the corresponding angles \( \theta \) in the interval \( [0^{\circ}, 360^{\circ}) \) by using the inverse cosine function: \( \theta = \cos^{-1}(x) \). Remember that cosine is positive in Quadrants I and IV, and negative in Quadrants II and III, so find all angles that satisfy each \( x \).
Write the solutions as exact values if possible, or round to the nearest tenth of a degree as appropriate, ensuring all solutions lie within the given interval.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Quadratic Equations in Trigonometric Functions

This concept involves treating trigonometric equations like algebraic quadratics by substituting the trigonometric function (e.g., cos θ) as a variable. Solving the quadratic yields possible values for the function, which can then be used to find the angle θ.
Recommended video:
5:35
Introduction to Quadratic Equations

Unit Circle and Angle Measurement

Understanding the unit circle helps relate cosine values to specific angles within the interval [0°, 360°). Each cosine value corresponds to one or two angles in this range, depending on the sign and magnitude, which is essential for finding all solutions.
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Introduction to the Unit Circle

Exact Values and Approximation in Trigonometry

Some trigonometric solutions correspond to well-known exact values (like cos 60° = 0.5), while others require approximation using a calculator. Recognizing when to use exact values or round to the nearest tenth degree ensures precise and appropriate answers.
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Fundamental Trigonometric Identities
Related Practice
Textbook Question

Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth of a degree, as appropriate.

sin 2θ = cos 2θ +1

Textbook Question

Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth of a degree, as appropriate.

sin² θ + 3 sin θ + 2 = 0

Textbook Question

Use the unit circle shown here to solve each simple trigonometric equation. If the variable is x, then solve over [0, 2π). If the variable is θ, then solve over [0°, 360°).                     

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cos θ = ―1/2

Textbook Question

Find the exact value of each real number y. Do not use a calculator.

y = cos⁻¹ (―√2/2)

Textbook Question

Solve each equation over the interval [0, 2π). Write solutions as exact values or to four decimal places, as appropriate.

tan² 2x -1 = 0

Textbook Question

Find the exact value of each real number y. Do not use a calculator.

y = tan⁻¹ (―√3)