Getting to the Roots of a Problem

Nevin Katz
6 min readDec 10, 2022


Image by Gerd Altmann from Pixabay

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.


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.


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.

A typical sin graph.

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 unit circle showing a value of 2π/3.

A few things:

  1. 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.


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!



Nevin Katz

Developer at EDC. I write about web development and biology. Subscribe at for article roundups.