Ch. 6 - Inverse Circular Functions and Trigonometric Equations

Lial12th EditionTrigonometryISBN: 9780136552161Not the one you use?Change textbook

Lial 12th Edition

Ch. 6 - Inverse Circular Functions and Trigonometric Equations

Problem 6.40

Chapter 7, Problem 6.40

Solve each equation for all exact solutions, in degrees.
2√3 cos (θ/2) = -3

Verified step by step guidance

Start by isolating the cosine term in the equation: \(2\sqrt{3} \cos\left(\frac{\theta}{2}\right) = -3\). Divide both sides by \(2\sqrt{3}\) to get \(\cos\left(\frac{\theta}{2}\right) = \frac{-3}{2\sqrt{3}}\).

Simplify the right-hand side by rationalizing the denominator if needed. This will give you a simplified exact value for \(\cos\left(\frac{\theta}{2}\right)\).

Use the inverse cosine function to find the principal value(s) of \(\frac{\theta}{2}\): \(\frac{\theta}{2} = \cos^{-1}(\text{value})\). Remember that cosine is positive in the first and fourth quadrants and negative in the second and third quadrants, so consider all angles where cosine equals this value.

Write the general solutions for \(\frac{\theta}{2}\) using the cosine periodicity: \(\frac{\theta}{2} = 360^\circ k \pm \alpha\), where \(\alpha\) is the reference angle found from the inverse cosine and \(k\) is any integer.

Finally, multiply all parts of the equation by 2 to solve for \(\theta\): \(\theta = 2 \times (360^\circ k \pm \alpha)\). This gives all exact solutions for \(\theta\) in degrees.

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Key Concepts

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

Solving Trigonometric Equations

This involves isolating the trigonometric function and finding all angle values that satisfy the equation within the given domain. For cosine equations, solutions often come in pairs due to the function's periodicity and symmetry.

Inverse Trigonometric Functions

Inverse cosine (arccos) is used to find the principal angle whose cosine equals a given value. Since cosine is positive in the first and fourth quadrants and negative in the second and third, understanding how to find all solutions requires considering these quadrants.

Angle Transformation and Domain Considerations

When the variable is inside a function argument like θ/2, solving requires adjusting the solution by multiplying or dividing accordingly. Additionally, since the problem asks for all solutions in degrees, one must consider the periodicity of cosine (360°) and adjust solutions to cover all possible angles.

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Related Practice

Textbook Question

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

y = sin⁻¹ √2/2

Textbook Question

Which one of the following equations has solution π?

a. arccos (―1) = x

b. arccos 1 = x

c. arcsin (―1) = x

Textbook Question

Decide whether each statement is true or false. If false, explain why.

The tangent and secant functions are undefined for the same values.

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

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

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.

√2 cos 2θ = -1

Solve each equation for all exact solutions, in degrees.2√3 cos (θ/2) = -3

Verified video answer for a similar problem:

Key Concepts

Solving Trigonometric Equations

Inverse Trigonometric Functions

Angle Transformation and Domain Considerations

Solve each equation for all exact solutions, in degrees.
2√3 cos (θ/2) = -3