Textbook Question
Find the exact value of each real number y. Do not use a calculator.
y = sin⁻¹ √2/2
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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 6.2.35
Lial 12th EditionCh. 6 - Inverse Circular Functions and Trigonometric EquationsProblem 6.2.35Chapter 7, Problem 6.2.35
Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth, as appropriate.
2 tan θ sin θ - tan θ = 0
Verified step by step guidance1
Start by writing down the given equation: \(2 \tan \theta \sin \theta - \tan \theta = 0\).
Factor out the common term \(\tan \theta\) from the left side: \(\tan \theta (2 \sin \theta - 1) = 0\).
Set each factor equal to zero to find possible solutions: \(\tan \theta = 0\) and \(2 \sin \theta - 1 = 0\).
Solve \(\tan \theta = 0\) by finding all angles \(\theta\) in \([0^\circ, 360^\circ)\) where the tangent function is zero.
Solve \(2 \sin \theta - 1 = 0\) by isolating \(\sin \theta\) to get \(\sin \theta = \frac{1}{2}\), then find all angles \(\theta\) in \([0^\circ, 360^\circ)\) that satisfy this.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Trigonometric Equations
Trigonometric equations involve functions like sine, cosine, and tangent. Solving these requires isolating the trigonometric function and finding all angle values within a given interval that satisfy the equation.
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How to Solve Linear Trigonometric Equations
Factoring and Zero-Product Property
Factoring expressions allows breaking down complex equations into simpler products. The zero-product property states that if a product equals zero, at least one factor must be zero, enabling separate equations to be solved.
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6:08Factoring
Solving for Angles in a Given Interval
When solving trigonometric equations over [0°, 360°), it is essential to find all angle solutions within one full rotation. This involves using reference angles and considering the signs of trig functions in each quadrant.
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2:28Solving Problems with Complementary & Supplementary Angles
Related Practice
Textbook Question
Which one of the following equations has solution π?
a. arccos (―1) = x
b. arccos 1 = x
c. arcsin (―1) = x
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Textbook Question
Solve each equation for x.
y = 1/2 tan (3x + 2), for x in [-2/3 - π/6, -2/3 + π/6]
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Textbook Question
Solve each equation over the interval [0°, 360°). Write solutions as exact values or to the nearest tenth, as appropriate.
sec² θ tan θ = 2 tan θ
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Textbook Question
Solve each equation in x over the interval [0, 2π) and each equation in θ over the interval [0°, 360°). Give exact solutions.
8 sec² x/2 = 4
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Textbook Question
The following equations cannot be solved by algebraic methods. Use a graphing calculator to find all solutions over the interval [0, 2π). Express solutions to four decimal places.
x² + sin x - x³ - cos x = 0
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