SWBAT use factoring and the reciprocal properties to solve trigonometric equations.

Why do these trig equations have so many solutions?

20 minutes

Solving equations is a topic that students have tons of experience with. This process will be much different with trigonometric equations due to the sheer number of solutions that students will be getting. Because of this change, students may have difficulty with this process, even though the general concept is something they are very familiar with.

I will preface this lesson by explaining to students that we will be solving equations that involve trig functions. We have already done very simple ones in the last unit, so I will start by revisiting an equation like **cos(x) = ½**. I will lead a discussion as we think about what the x value has to be. Almost always, the first student I choose will say that 60° is the solution. I will then ask students if that is the only solution. Eventually we will think about how there are an infinite number of solutions.

One of the **most important standards** I will establish is that we should always add multiples of the period to represent *all* solutions. So the answer to cos(x) = ½ should be x = 60° + 360°n or x = -60° + 360°n. This is really important because when we get an equation like cos(2x+ 4°) = ½, we can set up 60° + 360°n and then just subtract 4° and divide by 2 to get the answer. This format will keep us organized.

I start by giving students the worksheet with the three example problems and letting them get through as much as they can with their tables. Ten to fifteen minutes is usually enough for them to get some progress and have something to contribute to our class discussion.

25 minutes

Once it is time to share, I will choose students who have begun each equation correctly. This may not mean that they have the correct answer, but they at least were performing the correct algebraic steps. For example, I will choose a student who isolated sin^{2}x for the first equation or one who knew to factor the third equation.

After that, I will lead the discussion for using inverses to find the value of the argument in each equation. It is important to think out loud and **demonstrate the thought process** to find all solutions. For instance:

**My calculator gave me 57.7° as where the sine ratio is equal to sqrt(5/7). Sine is also positive in quadrant II, so I subtract 57.7° from 180° to get the other solution.**

Finding other solutions that your calculator does not give you is a big idea of this unit, so students are going to need to see how to do this.

Using a graph to analyze the solution set is also very important. I will go through this strategy for the first equation and use Desmos to show the solution set. With the graph, it is clear why there will be a new solution every 180°.

I can’t stress enough how important it is to** stay organized** while completing these equations. When students are getting the wrong answer, I often find that they are trying to do too much in their calculator or in their head and are missing solutions. Here is how I would go through the work for the second equation. You can specifically see the change from using cotangent to tangent, and then how I algebraically subtracted 7° after I set up an expression with multiples of the period.

10 minutes

At the end of this lesson, I really want to drive home the fact that there are an infinite number of solutions to these equations. I will ask "why do all of these equations have so many solutions?" to evoke responses. We can compare sin(x) = 1/2 to 3x + 5 = 19 to really hone in on the periodicity compared to the linearity. This is a **big idea** of trigonometric functions, so I want to make sure that students understand this. I discuss this closing in this video.

At this point, I will usually assign a few homework problems from the textbook in order to give some practice with solving equations.