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Chapter 5: Problem 2

Fill in the blanks. The equation \(2 \sin \theta+1=0\) has the solutions \(\theta=\frac{7 \pi}{6}+2 n \pi\) and \(\theta=\frac{11 \pi}{6}+2 n \pi,\) which are called _____ solutions.

Short Answer

Expert verified
The solutions are known as periodic solutions. The term 'periodic' is used due to the periodicity in the sine function, which causes the solutions to repeat after every \(2\pi\) radians.

Step by step solution

01

Understand the problem

The problem asks to find the type of solutions that \(\theta=\frac{7 \pi}{6}+2 n \pi\) and \(\theta=\frac{11 \pi}{6}+2 n \pi\) represent. It is a trigonometric equation where \(n\) represents any integer. Because sine function has a period of \(2\pi\), the solutions are \(2\pi\) apart.
02

Solve the equation

We are given the equation \(2 \sin \theta+1=0\). We rearrange the equation to find \(\sin \theta\) as follows: \(\sin \theta = -\frac{1}{2}\). The solutions to this equation are \(\theta=\frac{7 \pi}{6}+2 n \pi\) and \(\theta=\frac{11 \pi}{6}+2 n \pi\), which represent the angles for which the sine function equals -1/2 in the unit circle. The solutions repeat every \(2\pi\), the period of the sine function.
03

Determine the type of solutions

The solutions are \(2\pi\) apart, thus they are called periodic solutions, due to the periodicity in the sine function.

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Most popular questions from this chapter

Write the expression as the sine, cosine, or tangent of an angle. $$\frac{\tan 140^{\circ}-\tan 60^{\circ}}{1+\tan 140^{\circ} \tan 60^{\circ}}$$

Find the exact values of the sine, cosine, and tangent of the angle. $$-\frac{13 \pi}{12}$$

Find the exact values of the sine, cosine, and tangent of the angle. $$285^{\circ}$$

(a) use a graphing utility to graph the function and approximate the maximum and minimum points on the graph in the interval \([0,2 \pi),\) and (b) solve the trigonometric equation and demonstrate that its solutions are the \(x\) -coordinates of the maximum and minimum points of \(f .\) (Calculus is required to find the trigonometric equation.) Function $$f(x)=\sec x+\tan x-x$$ Trigonometric Equation $$\sec x \tan x+\sec ^{2} x-1=0$$

Use inverse functions where needed to find all solutions of the equation in the interval \([0,2 \pi)\). $$\tan ^{2} x+\tan x-12=0$$
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