Textbook Reference: Section 7.5 Solving Trigonometric Equations
At this point, we're going to start jumping around a bit and doing things in a slightly different order than the textbook, to make the material "flow" a bit better. If you want to get practice by doing exercises, try to look for exercises that look like what we're doing in and outside of class; if something looks unfamiliar, it might be something we'll get to later in the semester.
Ahh, solving equations ... the bread and butter of algebra. Brings back good memories, doesn't it?
Well ... it brings back memories, at least. Sadly, how good those memories are might vary.
In previous classes, you likely learned a whole ton of different techniques for solving equations, with different steps for different kinds of functions — linear functions, quadratic functions, polynomial functions, exponential functions, logarithmic functions, and so on. It might have felt overwhelming at times.
But I'm going to let you in on a little secret.
It turns out that there are really only two algebraic techniques for solving equations:
"Undoing" and "Splitting".
If you had me last semester, you already know this, and you can skip this section if you'd like. If not, or if you just want a quick refresher, read on.
Undoing and splitting
Suppose you I tell you I'm thinking of a number.
I can take my number, and do the following things in order:
- Cube it.
- Double it.
- Add seven.
- Divide it by five.
When I do these things, the answer I end up with is \(27\).
How can you figure out my original number?
You could just guess a bunch of different numbers and check whether you end up with \(27\). But that might take a while, so a more efficient and clever solution would be to roll the clock back and "undo" each of my steps:
- Start off with the answer, \(27\).
- Multiply it by \(5\), which gives \(135\).
- Subtract \(7\), which gives \(128\).
- Cut it in half, which gives \(64\).
- Finally, take the cube root, which gives \(4\).
Phrased algebraically, what you just did might look like this: \[ \begin{align*} \dfrac{2x^3+7}{5} &= 27\\ 2x^3+7 &= 135 & \text{Multiply both sides by }5\\ 2x^3 &= 128 & \text{Subtract }7\text{ from both sides}\\ x^3 &= 64 & \text{Divide both sides by }2\\ x &= 4 & \text{Take cube root of both sides} \end{align*} \]
What we're essentially doing is using the idea of inverse functions to get our variable by itself — each thing we did to \(x\) is like a layer, and by "undoing" those things, we're peeling back the layers.
We can run into some problems though. Take a look at this equation now: \[2|x+3|+1 = 7\] We can subtract \(1\) from both sides and then divide both sides by \(2\), but then we're left with: \[|x+3|=3\] It's more difficult to "peel back" the absolute value function, since it's not one-to-one — there are multiple inputs that could give us the same output.
But if we think about how absolute values work, we can realize that there are two numbers with an absolute value of \(3\): namely, \(3\) and \(-3\). This allows us to "split" our complicated equation into two easier-to-solve equations: \[x+3=3\qquad\qquad x+3=-3\] Solving these equations individually gives us our two solutions: \(x=0\) or \(x=-6\).
Using circular functions
Now that you're familiar with the unit circle, you already can solve some basic equations involving the circular functions.
...wait you are familiar with the unit circle, right?
If you haven't spent some time with your best-friend-the-unit-circle in some time, you might need to go back and review it. Sit down with it. Have some deep conversations. Get to where you can visualize the angle measures (in both degrees and radians, as well as the height, overness, and slope of those important points on the rim. Doing so will mean you spend less time going back and forth looking things up, and more time actually solving problems.
Consider the following equation: \[\sin(x)=\dfrac12\] What this is asking is, "what are all angles on the unit circle that have a height of \(\dfrac12\)?
If you look back at your unit circle — or, even better, if you can see it in your mind's eye — you should see that there are two solutions: \(x=\dfrac\pi6\), and \(x=\dfrac{5\pi}6\).
But is that all the solutions?
Remember, the circular functions are periodic, so they repeat forever for infinitely many coterminal angles. So, for example, \(x=\dfrac{13\pi}{6}\) would be a solution, as would \(x=-\dfrac{7\pi}{6}\).
We can see this even more clearly if we graph both sides of our equation:
Click the points of intersection to see their coordinates.
Each \(x\)-value where the two graphs intersect is a solution. (As you can see, there's quite a few.)
But I have some good news:
Once you can visualize angles on the unit circle, everything else is just applying what you already know about solving equations. So, most of the time, all you really have to do is attempt to get your circular function by itself (through one means or another), and then call up your best friend the unit circle to help you out!
Preview Activity 8
Answer these questions and submit your answers as a document on Moodle. (Please submit as .docx or .pdf if possible.)
- Find both solutions to the equation \(\tan(\theta)=-1\) in the interval \(0\le\theta\le 2\pi\). Then find at least two more solutions outside this interval. Include a unit circle sketch to support your answer.
- Solve the equation \(2\cos(x)+1=0\). Explain the steps you take to solve it. Include a unit circle sketch to support your answer.
- Come up with an equation (using a circular function) that has only one solution on the interval \([0,2\pi]\). Explain why it only has one solution. Include a unit circle sketch to support your answer.
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Answer AT LEAST one of the following questions:
- What was something you found interesting about this reading?
- What was an "a-ha" moment you had while doing this reading?
- What was the muddiest point of this reading for you?
- What question(s) do you have about anything you've read?
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