Problem 55 - Precalculus: A Concise Course

Question

Use a graph to solve the equation on the interval \([-2 \pi, 2 \pi]\). $$ \csc x=\sqrt{2} $$

Step-by-Step Solution

Answer

\[x = -\frac{7\pi }{4}, -\frac{\pi }{4}, \frac{\pi }{4}, \frac{3\pi }{4}, \frac{5\pi }{4}, \frac{7\pi }{4}\]

Step 1: Convert the Equation

Given equation is \(csc(x)=\sqrt{2}\). Convert the equation to sine form. So, the new equation becomes \(sin(x)=\frac{1}{\sqrt{2}}\)

Step 2: Determine the Quadrants

Since we know that sine is positive in the first and second quadrants, we can find the base angle in the first quadrant where \(sin(x)=\frac{1}{\sqrt{2}}\). This is \(\frac{\pi }{4}\). So the values of x in the first and second quadrant are \(\frac{\pi }{4}\) and \(\frac{3\pi }{4}\) respectively.

Step 3: Find More Solutions within the Interval

Solutions repeat after every period of 2π for sine function. So, other solutions within the given interval can be found by adding and subtracting multiples of 2π to the already found solutions. Our solutions become \(\frac{\pi }{4} +2k\pi \), \(\frac{3\pi }{4}+2k\pi \), where k is an integer.

Step 4: Choosing the Solutions

By putting different k values, we can get all solutions within the interval [-2π, 2π]. From these solutions we discard the ones outside the given interval. Here, the solutions are \[-\frac{7\pi }{4}, -\frac{\pi }{4}, \frac{\pi }{4}, \frac{3\pi }{4}, \frac{5\pi }{4}, \frac{7\pi }{4}\].

Key Concepts

Sine Function

The sine function is a fundamental concept in trigonometry, representing the y-coordinate of a point on the unit circle. When discussing the sine function, it's important to remember:

• It is periodic, meaning it repeats its behavior in regular intervals, specifically every \(2\pi\) radians.
• The range of sine values is between -1 and 1.
• The sine of an angle can be positive or negative depending on the quadrant it is in.

For an equation like \(\csc x = \sqrt{2}\), converting it to \(\sin x = \frac{1}{\sqrt{2}}\) allows us to pinpoint precise values of \(x\) that satisfy the condition. This sine value corresponds to specific angles on the unit circle.

Cosecant Function

The cosecant function, denoted as \(\csc(x)\), is the reciprocal of the sine function. It is defined as \(\csc(x) = \frac{1}{\sin(x)}\).

• Because it is the reciprocal, the cosecant function is undefined anywhere the sine function is zero.
• The range of \(\csc(x)\) excludes the interval between -1 and 1, as it only includes values outside this range.
• For \(\csc(x) = \sqrt{2}\), the corresponding \(\sin(x)\) value becomes \(\frac{1}{\sqrt{2}}\).

Understanding this function helps in solving equations that involve the sine's reciprocal form.

Unit Circle

The unit circle is a powerful tool in trigonometry. It's a circle with a radius of 1, centered at the origin of a coordinate plane. This circle helps to visualize the angles and their sine and cosine values.

• The circumference of the unit circle is \(2\pi\) radians.
• Each point on this circle corresponds to an angle, measured in radians or degrees.
• The x-coordinate is the cosine of the angle, while the y-coordinate is the sine of the angle.
• For \(\sin x = \frac{1}{\sqrt{2}}\), this corresponds to angles like \(\frac{\pi}{4}\) and \(\frac{3\pi}{4}\) on the unit circle.

Graphing on the unit circle provides a visual representation of the sine and other trigonometric functions.

Quadrants

Quadrants divide the coordinate plane into four sections, each indicating a different range of angles.

• Quadrant I: Both sine and cosine are positive.
• Quadrant II: Sine is positive, but cosine is negative.
• Quadrant III: Both sine and cosine are negative.
• Quadrant IV: Sine is negative, but cosine is positive.

Knowing which quadrant an angle is in is crucial. For \(\sin x = \frac{1}{\sqrt{2}}\), sine is positive, which means we focus on Quadrants I and II, yielding angles like \(\frac{\pi}{4}\) and \(\frac{3\pi}{4}\).