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 guidance

1. 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 \( \frac{1}{\sin^2 x} + \frac{1}{\sin x} - 2 = 0 \).
2. To simplify, multiply the entire equation by \( \sin^2 x \) (noting that \( \sin x \neq 0 \) in the domain) to clear the denominators: \( 1 + \sin x - 2 \sin^2 x = 0 \).
3. 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 \( 2 \sin^2 x - \sin x - 1 = 0 \).
4. 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 \( c=-1 \).
5. 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 \( \sin x = y \).

Key Concepts

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.

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.

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.