Textbook Question
In Exercises 85–96, use a calculator to solve each equation, correct to four decimal places, on the interval [0, 2𝝅). tan x = ﹣3
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5. Inverse Trigonometric Functions and Basic Trigonometric Equations
Linear Trigonometric Equations
6:09 minutesProblem 127
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. csc² x + csc x - 2 = 0
Verified step by step guidance1
Start by rewriting the given equation in terms of sine, since \( \csc x = \frac{1}{\sin x} \). The equation is \( \csc^2 x + \csc x - 2 = 0 \), so substitute to get \( \frac{1}{\sin^2 x} + \frac{1}{\sin x} - 2 = 0 \).
To simplify, multiply the entire equation by \( \sin^2 x \) (noting that \( \sin x \neq 0 \) in the domain) to clear the denominators: \( 1 + \sin x - 2 \sin^2 x = 0 \).
Rewrite the equation in standard quadratic form in terms of \( \sin x \): \( -2 \sin^2 x + \sin x + 1 = 0 \). For easier handling, multiply both sides by \( -1 \) to get \( 2 \sin^2 x - \sin x - 1 = 0 \).
Let \( y = \sin x \). Now solve the quadratic equation \( 2y^2 - y - 1 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=2 \), \( b=-1 \), and \( c=-1 \).
After finding the values of \( y \), determine which solutions lie within the range \( [-1, 1] \) since \( \sin x \) must be in this interval. Then, find all \( x \) in \( [0, 2\pi) \) such that \( \sin x = y \).
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6mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Reciprocal Trigonometric Functions
The cosecant function (csc x) is the reciprocal of sine, defined as csc x = 1/sin x. Understanding this relationship allows rewriting equations involving csc x in terms of sin x, which is often easier to solve.
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Solving Quadratic Equations in Trigonometric Form
Equations like csc² x + csc x - 2 = 0 can be treated as quadratic equations by substituting a trigonometric function (e.g., csc x = y). Solving the quadratic for y and then reverting to the trigonometric function helps find the solutions.
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Solving Trigonometric Equations on a Specific Interval
When solving trigonometric equations on [0, 2π), it is important to find all solutions within one full cycle of the unit circle. This involves considering the periodicity and domain restrictions of sine and cosecant functions.
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