Hi Everyone!
On this page you will find some material about Lesson 31. Read through the material below, watch the videos, and follow up with your instructor if you have questions.
Lesson 31: 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 31 (based on Lesson 31 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.
Warmup Question 1
Solve $\cos x =0$ for $x$ in $[0,2\pi)$.
Show Answer 1
$x=\dfrac{\pi}{2}, \dfrac{3\pi}{2}$
Warmup Question 2
Solve $\sin x =-1$ for $x$ in $[0,360^\circ)$.
Show Answer 2
$x=270^\circ$
Warmup Question 3
State the quadrant of the terminal side of $\theta$ given that $\tan \theta<0$ and $\sec \theta>0$.
Show Answer 3
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.
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 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.
Try Question 1
Solve $2\cos x +1=0$ with $x$ in $[0,2\pi)$.
Show Answer 1
$$2\cos x +1=0$$
$$2\cos x = -1$$
$$\cos x =-\dfrac{1}{2}$$
$$x=\arccos\left(-\dfrac{1}{2}\right)$$
$$x=\dfrac{2\pi}{3}$$
$x=\dfrac{2\pi}{3}$ 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 $\left\{\dfrac{2\pi}{3},\dfrac{4\pi}{3}\right\}$.
Try Question 2
Find all solutions in $[0,2\pi)$. State the solutions in radians using the exact form.
\[2\sin^2x-3\sin x +1 =0\]
Show Answer 2
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.
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!
Exit Questions
- 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 $[0, 360^{\circ})$.
Show Answer
$\bullet$ 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.
