Hi Everyone!

On this page you will find some material about Lesson 23. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

**Lesson 23: Trigonometric Equations**

Table of Contents

### Resources

*In this section you will find some important information about the specific resources related to this lesson: *

*the learning outcomes,**the section in the textbook,**the WeBWorK homework sets,**a link to the pdf of the lesson notes,**a link to a video lesson.*

**Learning Outcomes.** (from Coburn and Herdlick’s Trigonometry book)

- Use a graph to gain information about principal roots, roots in $[0, 2\pi$), and roots in $\mathbb R$.
- Use inverse functions to solve trig equations for the principal root.
- Solve trig equations for roots in $[0, 2\pi)$ or $[0^\circ, 360^\circ)$.
- Solve trig equations for roots in $\mathbb R$.
- Solve trig equations using fundamental identities.
- Solve trig equations using graphing technology.

**Topic**. This lesson covers

Section 6.3: Solving Basic Trigonometric Equations.

**WeBWorK**. There is one WeBWorK assignment on today’s material:

TrigEquations

**Lesson Notes.**

**Video Lesson.**

Video Lesson 23 (based on Lesson 23 Notes)

### Warmup Questions

*These are questions on fundamental concepts that you need to know before you can embark on this lesson. Don’t skip them! Take your time to do them, and check your answer by clicking on the “Show Answer” tab.*

Solve $\cos x =0$ for $x$ in $[0,2\pi)$.

$x=\dfrac{\pi}{2}, \dfrac{3\pi}{2}$

Solve $\sin x =-1$ for $x$ in $[0,360^\circ)$.

$x=270^\circ$

State the quadrant of the terminal side of $\theta$ given that $\tan \theta<0$ and $\sec \theta>0$.

Since $\sec\theta>0$, we have $\cos\theta>0$, so the terminal side of $\theta$ must lie in QI or QIV. Since $\tan\theta<0$, the terminal side of $\theta$ must lie in QII or QIV. Since both conditions must be true, we conclude that the quadrant of the terminal side of $\theta$ is QIV.

### Review

*If you are not comfortable with the Warmup Questions, don’t give up! Click on the indicated lesson for a quick catchup. A brief review will help you boost your confidence to start the new lesson, and that’s perfectly fine.*

Need a review? Check Lesson 19 and Lesson 20.

### Quick Intro

*This is like a mini-lesson with an overview of the main objects of study. It will often contain a list of key words, definitions and properties – all that is new in this lesson. We will use this opportunity to make connections with other concepts. It can be also used as a review of the lesson.*

**A Quick Intro to Trigonometric Equations**

**Key Words.** Inverse trigonometric expressions, trigonometric equations, symmetry

$\bigstar$ In the Warmup Question 1, we saw that $\cos x = 0 $ when $x=\dfrac{\pi}{2}, \dfrac{3\pi}{2}$. Equivalently, we can say

$$\arccos 0 = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$$

or

$$\cos^{-1}0 =\dfrac{\pi}{2}, \dfrac{3\pi}{2}$$

to represent the angles whose cosine is zero between 0 and $2\pi$.

$\bigstar$ The main idea in solving a trigonometric equation is to reduce the equation so that there is a trigonometric expression such as $\cos$, $\sin$ and $\tan$ on one side, and a number on the other side.

$\bigstar$ In the above example, there are two solutions between 0 and $2\pi$. The calculator only provides one $\arccos$ or $\cos^{-1}$ value. To determine the other solution, it is useful to remember that on the unit circle, $x$ represents the cosine value and $y$ represents the sine value. For instance, if $\cos\theta$ is negative, then $x$ is negative, which happens when the terminal side of $\theta$ is in QII or QIII. You can then use the symmetry discussed in Lesson 20 to determine all solutions.

### Video Lesson

*Many times the mini-lesson will not be enough for you to start working on the problems. You need to see someone explaining the material to you. In the video you will find a variety of examples, solved step-by-step – starting from a simple one to a more complex one. Feel free to play them as many times as you need. Pause, rewind, replay, stop… follow your pace!*

**Video Lesson**

**A description of the video**

In the video you will see how to solve

- $\cos u = -\dfrac{1}{2}$, $u\in [0,2\pi)$
- $\tan u = -1$, $u\in[0,2\pi)$
- $\sin u = -\dfrac{3}{5}$, $u\in [0,2\pi)$

### Try Questions

*Now that you have read the material and watched the video, it is your turn to put in practice what you have learned. We encourage you to try the Try Questions on your own. When you are done, click on the “Show answer” tab to see if you got the correct answer.*

Solve $2\cos x +1=0$ with $x$ in $[0,2\pi)$.

$$2\cos x +1=0$$

$$2\cos x = -1$$

$$\cos x =-\dfrac{1}{2}$$

$$x=\arccos\left(-\dfrac{1}{2}\right)$$

$$x=\dfrac{2\pi}{3}$$

$x=\dfrac{2\pi}{3}$ is in QII

Since $\cos x<0$, there is another solution in QIII, which can be found by using symmetry:

$$x=\dfrac{3\pi}{2}-\dfrac{\pi}{6} = \dfrac{9\pi}{6}-\dfrac{\pi}{6}=\dfrac{8\pi}{6}=\dfrac{4\pi}{3}.$$

The solution set is $\left\{\dfrac{2\pi}{3},\dfrac{4\pi}{3}\right\}$.

Find all solutions in $[0,2\pi)$. State the solutions in radians using the exact form.

\[2\sin^2x-3\sin x +1 =0\]

Let $u=\sin x$. Then

\[2\sin^2x-3\sin x +1 =0 \;\Longrightarrow\; 2u^2-3u+1=0 \;\Longrightarrow\; (2u-1)(u-1)=0\]

\[ \;\Longrightarrow\; 2u-1=0 \text{ or } u-1=0 \;\Longrightarrow\; u=\dfrac{1}{2} \text{ or } u=1\]

When $u=\dfrac{1}{2}$, we have $\sin x = \dfrac{1}{2} \;\Longrightarrow\; x=\dfrac{\pi}{6} \text{ or } \pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}$.

When $u=1$, we have $\sin x = 1 \;\Longrightarrow\; x=\dfrac{\pi}{2}$.

The solution set is $\left\{\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{\pi}{2}\right\}$.

### WeBWorK

*You should now be ready to start working on the WeBWorK problems. Doing the homework is an essential part of learning. It will help you practice the lesson and reinforce your knowledge.*

**WeBWork**

It is time to do the homework on WeBWork:

TrigEquations

When you are done, come back to this page for the Exit Questions.

### A STEM Application: Alternating Circuits

Check it here

### Exit Questions

*After doing the WeBWorK problems, come back to this page. The Exit Questions include vocabulary checking and conceptual questions. Knowing the vocabulary accurately is important for us to communicate. You will also find one last problem. All these questions will give you an idea as to whether or not you have mastered the material. Remember: the “Show Answer” tab is there for you to check your work!*

- How can we solve $\cos(x)=0.3$, $x \in [0,2\pi)$? How do we find all real solutions?
- How do we see this as a system of equations and represent these as $x$-coordinates of points on the graph of $y=\cos(x)$?

$\bigstar$ Find all solutions of $3\sin x + 1 =0$ in $[0, 360^{\circ})$.

$\bullet$ We have $\sin x = -\dfrac{1}{3}$. So $x= \sin^{-1}\left(-\dfrac{1}{3}\right)$. Using a calculator, we find that $x = -19.47^{\circ}$. Since $x$ is negative, we need to find a coterminal angle between $0^{\circ}$ and $360^{\circ}$.

The coterminal angle is $360^{\circ} – 19.47^{\circ} = 340.53^{\circ}$. This is the solution in QIV. The reference angle is $19.47^{\circ}$.

$\bullet$ Since $\sin x <0$, there is another solution in QIII. The reference angle is the same: $19.47^{\circ}$.

So the other solution must be

$$180^{\circ}+ 19.47^{\circ} = 199.47^{\circ}.$$

$\bullet$ The solutions to the equation $3\sin x + 1 =0$ in $[0, 360^{\circ})$ are $199.47^{\circ}$ and $340.53^{\circ}$.

### Need more help?

*Don’t wait too long to do the following.*

- Watch the additional video resources.

- Talk to your instructor.
- Form a study group.
- Visit a tutor. For more information, check the tutoring page.