Textbook Question
In Exercises 85–96, use a calculator to solve each equation, correct to four decimal places, on the interval [0, 2𝝅). cos² x - cos x - 1 = 0
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5. Inverse Trigonometric Functions and Basic Trigonometric Equations
Linear Trigonometric Equations
5:47 minutesProblem 129
Textbook Question
In Exercises 127–130, solve each equation on the interval [0, 2𝝅) by first rewriting the equation in terms of sines or cosines. sec² x + 3 sec x + 2 = 0
Verified step by step guidance1
Recall that secant is the reciprocal of cosine, so rewrite the equation in terms of cosine: replace \( \sec x \) with \( \frac{1}{\cos x} \). The equation becomes \( \left( \frac{1}{\cos x} \right)^2 + 3 \left( \frac{1}{\cos x} \right) + 2 = 0 \).
Multiply through the entire equation by \( \cos^2 x \) to clear the denominators, resulting in a quadratic equation in terms of \( \cos x \): \( 1 + 3 \cos x + 2 \cos^2 x = 0 \).
Rewrite the quadratic equation in standard form: \( 2 \cos^2 x + 3 \cos x + 1 = 0 \).
Use the quadratic formula or factoring to solve for \( \cos x \). The quadratic formula is \( \cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=2 \), \( b=3 \), and \( c=1 \).
Find all values of \( x \) in the interval \( [0, 2\pi) \) such that \( \cos x \) equals the solutions found. Use the unit circle or inverse cosine function to determine these angles.
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Video duration:
5mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Secant Function and Its Relationship to Cosine
The secant function, sec x, is the reciprocal of the cosine function, defined as sec x = 1/cos x. Understanding this relationship allows rewriting equations involving sec x in terms of cosine, which is often easier to solve.
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Graphs of Secant and Cosecant Functions
Solving Quadratic Equations in Trigonometric Functions
Trigonometric equations like sec² x + 3 sec x + 2 = 0 can be treated as quadratic equations by substituting a trigonometric expression (e.g., sec x = y). Solving the quadratic yields values for the trigonometric function, which can then be used to find the angle solutions.
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Finding Solutions on a Specific Interval [0, 2π)
When solving trigonometric equations, it is important to find all solutions within the given interval, here [0, 2π). This involves considering the periodicity of sine and cosine functions and checking all possible angles that satisfy the equation within one full rotation.
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