# Getting to the Roots of a Problem

Recently I came across a vexing trigonometry problem that uses a number of different concepts, such as unit circles and factoring. I thought it would be helpful to share what I found while it’s fresh in my mind.

This one is Problem #4 from AnalyzeMath, a terrific site with free math tutorials. The problem is below.

Solve the trigonometric equation given by

sin(x) + sin(x/2) = 0 for 0 ≤ x ≤ 2π.

While the solution is further down, I found I had to break it down further to fully understand it. Below is what I came up with.

**Roadmap**

Whenever I work with my students, I try to encourage them to develop a basic plan for solving the problems. Here were my initial thoughts:

- The fact that the equation is set equal to zero makes me think there are multiple roots.
- That x/2 in sin(x/2) looks hard to work with.
- If I have to use x/2, I want it in all parts of the equation so I have just one input to work with.
- Given this, it would be nice to get that first expression, sin(x), in terms of x/2.
- Assuming we can get both expressions in terms of x/2, let’s solve for x using what you know about sinusoids and unit circles.

# Using the identity to achieve like terms

For the term, sin(x), let’s try the following identity.

sin(2x) = 2sin(x) • cos(x)

We want 2sin(x) in terms of x/2, but to use this identity, let’s get there indirectly. Let’s create a variable v, which we substitute for x/2.

Here’s the sin(x) in the original problem.

sin(x) + sin(x/2) = 0

Applying the identity and** v = x/2, **we can change sin(x) in the following way.

sin(x) = sin(2v)

Now let’s apply the identity.

sin(2v) = 2 sin(v) • cos(v)

How will it look with x/2 swapped back in? Let’s find out.

sin(2 • x/2) = 2 sin(x/2) cos(x/2)

We can also simply state this in terms of sin(x).

sin(x) = 2 sin(x/2) cos(x/2)

Replacing **sin(x) **with** 2 sin (x/2) cos (x/2)** into the original equation results in the following equation, with the new term on the left.

2sin(

x/2) cos(x/2) + sin(x/2) = 0

We now have like terms! All parts of the equation now use x/2.

# Factors

Now that sin(x/2) is in both terms, we have a possible factor in our sights: sin(x/2), which part of each term. Let’s prepare to factor it out by rearranging a few terms.

Let’s review where sin(x/2) is in each term.

We can now factor out sin(x/2), as shown below.

Nice! This gives us the two factors: **sin(x/2) **and **2cos(x/2) + 1.**

Since multiplying one factor by the other gives us zero, *both factors* should be equal to zero in the equation. In other words,

- sin(x/2) = 0
- 2cos(x/2)+1 = 0

With the equations in hand, let’s find the x-value(s) that make each equation true.

**Solving sin(x/2) = 0**

For the factor sin(x/2), the term x/2 is effectively our input in a sin graph.

We also know that in a regular y=sin(x) graph,

sin(0) = 0 and sin (π) = 0.

In other words, when the input value is 0 or π, the output, or y value, is 0.

So since x/2 is our input in sin(x/2) = 0, this means that x/2 can equal zero.

Because 2 • x/2 = 0, x is equal to zero as well.

Nice! We found the value of x for the first root. Nowm let’s find the x for the root that is π.

Since x/2 is our input, x/2 = π. Therefore,

x = 2 • x/2 = 2 • π = 2π.

So given that sin(x/2) = 0, we have two possible solutions: **2π and 0**. Now let’s examine the thornier factor, 2cos(x/2)+1, to see if there are any others.

**Solving 2cos(x/2)+1 = 0**

Let’s simplify this to make it easier to work with.

2cos(x/2) +1 = 0

Subtracting 1 from both sides yields this:

2cos(x/2) = -1

Then, dividing both sides by two yields this:

cos(x/2) = -½

Rather than make a graph this time, I found it easier to figure out what x/2 would be with a unit circle. Here’s the one I drew.

A few things:

- When the cosine is -½, the ratio of the adjacent side to hypotenuse is -½.

2. Because it’s negative, there is a possible solution in the upper-left quadrant.

3. Because the ratio of the adjacent side to longest side is 1:2, the reference angle — the angle between the red line and the nearest horizontal — will be 60 degrees, or π/3. (That is a pattern in 30–60–90 triangles.)

4. This brings us to a total angle of 2π/3, which is signified by the area in yellow.

5. This means that an input value of 2π/3, or 120 degrees, results in an output value of -½.

In other words, cos(2π/3) = -½

And if if cos(x/2) = -½ and, then x/2 = 2π/3.

With an input value of 2π/3, we can say this:

x = 2 • x/2 = 2 • 2π/3 = 4π/3.

Thus, the one root that comes out of the factor with cosine is **4π/3**.

**Quick note:** Looking back at step 2 above, a solution in the lower left quadrant may have seemed plausible since an input value of 4π/3, or 240 degrees, also results in an output of -1/2. However, the resulting x value would have been 8π/3, which is outside the bounds we were given at the outset, since x has to be between 0 and 2π.

# Summary

Let’s summarize the roots we get from each factor.

- sin(x/2) = 0 gives us x = 0 or 2π.
- 2cos(x/2)+1 = 0 give us x = 4π/3.

The equation with its roots is shown below, and you can view the original graph on desmos.

That is how we can find the roots using trigonometry.

## Sketching the graph

If I had to sketch the graph, these roots would help a lot, though I’d want to find the y-value of the point midway between the 0 and 4π/3, as well as the one midway between 4π/3 and 2π so I could represent the hill and valley with reasonable accuracy.

## Where we’ve been

Here is how we found the roots of* ***sin(x) + sin(x/2) = 0.**

- First, we used the identity
**sin(2x) = 2sin(x) • cos(x)**— as well as a little substitution — to make an alternate expression for sin(x) that used x/2. - Our resulting equation was
**2sin(x/2) cos(x/2) + sin(x/2) = 0**. - Factoring out sin(x/2) gave us
**sin(x/2) • (2cos(x/2) + 1) = 0**. - This gave us two factors: sin(x/2) and 2cos(x/2)+1, which we set equal to zero.
- Solving sin(x/2) = 0 gave us two roots: 0 and π.
- Solving 2cos(x/2)+1 = 0 gave us one additional root: 4π/3, or 240 degrees.

The most challenging part I found about this problem was using that first identity correctly. Before doing so, I had to realize that getting both terms to have x/2 would not overcomplicate it, but instead allow me to factor out sin(x/2).

I hope this has been helpful article on how to solve this problem. If you’d like me to break down other problems, let me know. Cheers!