Question

In Exercises \(31-36,\) solve the equation in the specified interval. $$\csc x=2\( \)0 < x <\( 2\)\pi$$

Short answer

The solutions for the equation \(\csc x = 2\) in the interval \(0 < x < 2\pi\) are \(x = \pi / 6\) and \(x = 5\pi / 6\).

Step-by-step solution

Understand the cosecant function

The cosecant function, \(\csc(x)\), is the reciprocal of the sine function. Thus, \(\csc(x) = 1 / \sin(x)\). Therefore, the equation \(\csc(x) = 2\) implies that \(\sin(x) = 1 / 2\). Because the sine function outputs values between -1 and +1, the equation has solutions.

Determine the sine inverses

To determine the \(x\)-values in the given interval \(0 < x < 2\pi\) that satisfy \(\sin(x) = 1 / 2\), use the unit circle or sine graph. Remember that \(\sin(x) = 1 / 2\) for \(x = \pi / 6\) and \(x = 5\pi / 6\). Both of these values fall in the interval \(0 < x < 2\pi\).

Conclusion

Therefore, the solutions to the equation \(\csc(x) = 2\) in the interval \(0 < x < 2\pi\) are \(x = \pi / 6\) and \(x = 5\pi / 6\).

Key concepts

Cosecant Function

The cosecant function, denoted as \(\csc(x)\), is an important trigonometric function in the study of right triangles and waves. It is defined as the reciprocal of the sine function, which means \(\csc(x) = \frac{1}{\sin(x)}\). Understanding the behavior of the cosecant function helps with solving trigonometric equations.

When an equation sets the cosecant function to a particular value, such as \(\csc(x) = 2\), it implies that the sine of the angle \(x\) is the reciprocal of that value, so \(\sin(x) = \frac{1}{2}\). It's noteworthy that the cosecant function, like all reciprocal trigonometric functions, can have values outside the range [-1,1], unlike the sine function which is confined to that range.

Another key point to remember is that the cosecant function tends to infinity as the sine function approaches zero, since division by a small number yields a large one. Therefore, when using the cosecant function in equations, we should consider all possible angles where the sine takes on the corresponding reciprocal value.

Unit Circle

The unit circle is a crucial tool in trigonometry for visualizing and understanding the relationship between angles and trigonometric functions. It's a circle with a radius of 1, centered at the origin of a coordinate plane.

Every point \( (x, y) \) on the circumference can be related to an angle \( \theta \) by using the sine and cosine functions: the x-coordinate is equivalent to \( \cos(\theta) \) and the y-coordinate to \( \sin(\theta) \). The sine and cosine values are the projections of the radius forming the angle onto the x- and y-axes, respectively.

When solving \( \sin(x) = \frac{1}{2} \) by relating it to the unit circle, you'll find that there are two angles where the y-value (sine value) is exactly 1/2 within a full rotation (0 to \(2\pi\)). These are at \( \pi/6 \) and \( 5\pi/6 \) radians, corresponding to 30 and 150 degrees. By understanding the symmetry and the periodic nature of the sine function in the unit circle, you can determine these critical angles easily.

Sine Inverse

The sine inverse function, also known as arcsine and denoted as \( \sin^{-1}(x) \) or \( \arcsin(x) \), is the inverse operation of the sine function. It's used to find the angle whose sine is a given number. This function is particularly useful when you are given the sine of an angle and need to determine the angle itself.

The output of the sine inverse function is an angle, typically measured in radians or degrees. However, because sine is not a one-to-one function (multiple angles can have the same sine value), the sine inverse is defined to return angles only within a certain range, usually \( -\pi/2 \) to \( \pi/2 \) for radians (-90 to 90 degrees).

When solving an equation like \( \sin(x) = \frac{1}{2} \) for \(x\), you often get multiple solutions due to the periodicity of the sine function. In the range of \( 0 < x < 2\pi \) specified in our textbook exercise, \( \sin^{-1}(\frac{1}{2}) \) gives \( \pi/6 \) directly, but considering sine's periodic nature and symmetry, you also get \( 5\pi/6 \) as a valid solution within the specified interval.