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
Ch. 6 - Inverse Circular Functions and Trigonometric Equations
Lial12th EditionTrigonometryISBN: 9780136552161
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Table of contents
- Ch. R - Algebra Review381
- Ch. 1 - Trigonometric Functions188
- Ch. 2 - Acute Angles and Right Triangles168
- Ch. 3 - Radian Measure and The Unit Circle155
- Ch. 4 - Graphs of the Circular Functions100
- Ch. 5 - Trigonometric Identities238
- Ch. 6 - Inverse Circular Functions and Trigonometric Equations158
- Ch. 7 - Applications of Trigonometry and Vectors148
- Ch. 8 - Complex Numbers, Polar Equations, and Parametric Equations15
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Lial 12th EditionCh. 6 - Inverse Circular Functions and Trigonometric EquationsProblem 6.RE.45
Lial 12th EditionCh. 6 - Inverse Circular Functions and Trigonometric EquationsProblem 6.RE.45Chapter 7, Problem 6.RE.45
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
Verified step by step guidance1
Recognize that the equation is a quadratic in terms of \(\sin \theta\). Rewrite the equation as \(\sin^{2} \theta + 3 \sin \theta + 2 = 0\) to emphasize this.
Let \(x = \sin \theta\). Substitute into the equation to get a quadratic equation: \(x^{2} + 3x + 2 = 0\).
Factor the quadratic equation: find two numbers that multiply to 2 and add to 3. This gives \((x + 1)(x + 2) = 0\).
Solve each factor for \(x\): \(x + 1 = 0\) gives \(x = -1\), and \(x + 2 = 0\) gives \(x = -2\). Since \(\sin \theta\) must be between -1 and 1, discard \(x = -2\).
Find all angles \(\theta\) in the interval \([0^\circ, 360^\circ)\) such that \(\sin \theta = -1\). Use the unit circle or inverse sine function to determine these values.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Solving Quadratic Equations in Trigonometric Form
The given equation sin²θ + 3 sinθ + 2 = 0 is a quadratic equation in terms of sinθ. To solve it, treat sinθ as a variable (e.g., x), solve the quadratic equation for x, and then find θ values that satisfy sinθ = x within the given interval.
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Solving Quadratic Equations by Completing the Square
Unit Circle and Sine Values
Understanding the unit circle is essential to find all angles θ where sinθ equals a specific value. Since sine corresponds to the y-coordinate on the unit circle, solutions for sinθ = x can be found by identifying angles in [0°, 360°) with that sine value.
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6:34Sine, Cosine, & Tangent on the Unit Circle
Interval Restriction and Solution Representation
The problem restricts θ to the interval [0°, 360°), so only solutions within one full rotation are valid. Solutions should be expressed as exact values (like 30°, 45°) or rounded to the nearest tenth of a degree, depending on whether the sine values correspond to standard angles.
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4:34How to Solve Linear Trigonometric Equations
Related Practice
Textbook Question
Solve each equation over the interval [0, 2π). Write solutions as exact values or to four decimal places, as appropriate.
tan x = cot x
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.
3 cos² θ + 2 cos θ - 1 = 0
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
Solve each equation over the interval [0, 2π). Write solutions as exact values or to four decimal places, as appropriate.
2 tan x -1 = 0
Textbook Question
Solve each equation for all exact solutions, in radians.
cos 2x + cos x = 0