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In Exercises 1-36, solve each of the trigonometric equations exactly on the interval \(0 \leq x<2 \pi\). $$ \sin x+\csc x=2 $$

Short Answer

Expert verified
The equation \( \sin x + \csc x = 2 \) is solved by \( x = \frac{\pi}{2} \) on the interval \(0 \leq x < 2\pi\).

Step by step solution

01

Express Cosecant in Terms of Sine

Given the equation \( \sin x + \csc x = 2 \), we first express cosecant in terms of sine. Recall that the cosecant function is the reciprocal of the sine function: \( \csc x = \frac{1}{\sin x} \). Substituting this in gives us: \( \sin x + \frac{1}{\sin x} = 2 \).
02

Multiply to Clear the Fraction

Multiply both sides of the equation by \( \sin x \) to eliminate the fraction: \( \sin^2 x + 1 = 2 \sin x \). This simplifies to the quadratic equation: \( \sin^2 x - 2 \sin x + 1 = 0 \).
03

Solve the Quadratic Equation

The equation \( \sin^2 x - 2 \sin x + 1 = 0 \) is a perfect square trinomial. It can be factored as \( (\sin x - 1)^2 = 0 \), which implies \( \sin x = 1 \).
04

Determine Solutions for \( x \)

Since \( \sin x = 1 \), we consider the interval \( 0 \leq x < 2\pi \). The solution is \( x = \frac{\pi}{2} \) because \( \sin \left(\frac{\pi}{2}\right) = 1 \) and it is the only angle in this interval where the sine function equals 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sine Function
The sine function, denoted as \( \sin x \), is one of the primary functions in trigonometry. It's defined as the ratio of the opposite side to the hypotenuse in a right triangle. For an angle \( x \), the sine function gives the vertical coordinate of the point on the unit circle corresponding to that angle. It is a periodic function with a period of \( 2\pi \), which means it repeats its values every \( 2\pi \) units. This property is helpful when solving trigonometric equations within specific intervals. The range of the sine function is between -1 and 1, making it useful in modeling wave-like phenomena.
  • Range: \([-1, 1]\)
  • Period: \(2\pi\)
  • Key Values: \( \sin 0 = 0 \), \( \sin \frac{\pi}{2} = 1\), \( \sin \pi = 0 \), and \( \sin \frac{3\pi}{2} = -1\)
It's important to detect when \( \sin x \) attains specific values, like 0, 1, or -1, to determine solutions to trigonometric equations on given intervals.
Cosecant Function
The cosecant function, denoted \( \csc x \), is the reciprocal of the sine function. It is defined as \( \csc x = \frac{1}{\sin x} \). Because it is a reciprocal, \( \csc x \) is undefined wherever \( \sin x = 0 \). This occurs at multiples of \( \pi \), where the sine function equals zero.The range of cosecant does not include values between -1 and 1 because these would require dividing by a number greater than 1 in absolute value, resulting in undefined or infinite values.
  • Domain: All real numbers except integer multiples of \( \pi \)
  • Range: \(( -\infty, -1 ] \cup [ 1, \infty ) \)
The concept of \( \csc x \) is useful in trigonometric equations involving reciprocals, like those in our exercise. By expressing \( \csc x \) as \( \frac{1}{\sin x} \), equations can be transformed to focus on \( \sin x \), making them simpler to solve.
Quadratic Equation
A quadratic equation is any equation that can be written in the form \( ax^2 + bx + c = 0 \). In trigonometry, we often encounter quadratic equations with trigonometric functions, as in the given problem.The equation \( \sin^2 x - 2 \sin x + 1 = 0 \) is a quadratic in terms of \( \sin x \). It can be treated like any standard quadratic equation.
  • Standard form: \( a(x - p)^2 = 0 \) for perfect squares
  • Solution method: Factoring, quadratic formula, or completing the square
For this equation, recognizing it as a perfect square \((\sin x - 1)^2 = 0\) makes it straightforward to solve, as \( \sin x = 1 \). Such simplifications allow us to find the values of \( x \) that satisfy the original trigonometric equation.
Interval Notation
Interval notation is a mathematical shorthand used to describe a range of numbers along an axis. It helps clearly define the set of solutions for an equation, especially when these solutions are limited to specific intervals.In the context of trigonometric equations, interval notation specifies where solutions are valid. For instance, the interval \( 0 \leq x < 2\pi \) indicates that \( x \) includes all values from 0 up to, but not including, \( 2\pi \). There are specific symbols used in interval notation:
  • Brackets \([\ ]\): Denote inclusion of endpoint (closed interval)
  • Parentheses \((\ )\): Denote exclusion of endpoint (open interval)
Understanding intervals is crucial in trigonometry as most periodic functions are evaluated within a set interval, necessitating precision in which endpoints are or aren't part of the solution set.

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Most popular questions from this chapter

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