SWBAT solve trigonometric equations.

How do you solve a trigonometric equation when the period is not 1?

10 minutes

Today students will continue to practice solving trigonometric equations. The new challenges posed today include equations period changes and horizontal shifts.

I begin the lesson by asking students to discuss how the answers will change when there is a coefficient in front of x. As they consider this prompt, some students will try to solve each problem. When they do, some will get stuck. Other students will employ a calculator to graph the equation and visually identify solutions. I will observe carefully to see which ideas come to my students' minds as they approach the problems. After a few minutes I will begin asking questions:

- Compare the equations that you are working on: how are they alike? how are they different?
- How does a coefficient affect the graph of a trig function?
- If the function is stretched horizontally over an interval, how will that affect the number of solutions to a trig equation?
- What happens if the function shrinks on the interval?

After we have had a brief conversation where students share their ideas about these questions, we will look at the graphs of each function. As a group we will determine that the first equation has two solutions, the middle equation has four solutions, and the final equation has one solution.

20 minutes

Next I will demonstrate how to solve the problems. Comparing the three equations, I will ask students:

**Is it correct to say that 2x and x/3 have to equal the same values as the first equation? Why?**

Some of my students will struggle with this abstract question. As questions arise, this is a good opportunity to discuss the use of substitution as a problem solving strategy. I define a variable, say w with a statement like w = 2x. Beneath I will write,

How can we find values for w that make this sentence true: **sin w=1/2** ?

Once they find w we replace w with 2x in our solution. This move helps move students towards on understanding of the idea that in the second and third examples, 2x and x/3 must equal the same values as the results we find for x in the first equation. Using this information, we quickly move to the situation where we **solve 2x = pi/6** or **2x = 5pi/6**. Most of my students can now correctly solver for the value of x.

Our next conversation returns us to the question of "how many solutions exist?" In a straight forward way I ask, "Have we identified all possible solutions?" If needed I put the graph of y=sin 2x on the board to show that there are, in fact, four solutions to be named. At this point I let my students use the graph to see the other solutions. Then, we can work backward to to see the relationship between these unknown solutions and the ones we found. I want students to see that you add 2*pi to the existing solutions to find the other two solutions.

We follow the same process for the last problem. When we check students realize that 5pi/2 is not in the interval so it is not a solution for this situation. Looking at the graph will also allow students to see that that the solutions is correct.

After working with these problems I ask students to consider this:

- How would you find all the solutions if the equation was sin 3x=1/2 or sin 4x=1/2.
- What is the period?
- How many solutions will it have?

I give students about three minutes to consider these questions. Students determine that they will add 2pi to the first answers to get the second set and continue adding 2pi until they get all the solutions.

5 minutes

As class ends I ask students to complete an Exit Slip. I will choose a problem that helps me to see how much my students understand of what was discussed today. In general, I expect that some students will struggle to find all of the solutions, so I will likely let students help each other as they work on this problem. Within this unit, I expect my students to rely on each other, as well as on me for assistance.