Here's an example, **sin(x) = √2/2**. We know that the **sin(π/2) = √2/2**, but so do **sin(3π/2)**, **sin(9π/2)**, **sin(11π/2)**, and so on ... see the graph below. The solutions to the equation are the intersections of the function **f(x) = sin(x)** and **g(x) = √2/2**.

In this section we'll do several examples of trig-equation solving, and eventually I'll put some videos up to help you. But first, there are a few tools you'll need to have at your disposal ...

It's just crucial for you to be able to draw a quick sketch – on paper or in your head – of one or more cycles of each of the three basic trig functions. That means knowing both the general shapes of the curves, that sine starts at zero and cosine starts at +1, and that the tangent is a repeating sigmoid (sideways S-shaped) curve.

Be able to jot down the common angles, in multiples of **π/4** and **π/6**. I've only included the **π/4** angles in these figures, you'll get a refresher on the others if you work through the examples below. Sketching these graphs is really just a matter of *counting*, much as you'd do for the unit circle. Start by writing **0** and **2π**, then divide that into chunks of **π**, then **π/2**, then divide each of those into **π/4** or **π/6 **chunks

To further refresh your memory, a cycle (plus a little more on each side) of the tangent function and the right triangles we most commonly use are shown below.

In this problem, we recognize that **√**3 is present in the 30-60-90 special triangle, so it makes sense to look there for some solutions. Here's that triangle:

Now notice that

Now here's a graph of **tan(x)** between **[-2π, 2π]**. On it, we'll plot our **initial solution**, **x = √3/3**

The red line is the line **y = √3**. Obviously, on **[-2π, 2π]** there are three more solutions, places where the tangent function is equal to **√3**. We need to find those. Fortunately, the tangent function has very regular behavior, which just reduces the rest of the problem to some simple arithmetic. Notice that every curve in the tangent graph is separated by **π** radians. That means our solutions, one per curve, are **π** rad apart. So we just need to add multiples of **π**, or better yet **6π/6** for the common denominator, like this:

So on our interval, **[-2π, 2π]**, we have four solutions to this trig equation. They are:

We found these by first finding one solution, the easiest, which we found by remembering the 30-60-90 triangle, then we recognized the periodicity of the tangent function – repeating every **π** radians – to walk through our interval and count out the rest of the solutions.

In this case, that's dividing by 2 on both sides:

Now hopefully that **√2/2 **will trigger the memory of the 45-45-90 triangle, and how we can use it to find trig functions for 45˚ angles.

So we have as our basic solution

Now let's sketch a graph of the sine function between **0** and **4π**. We'll sketch in the line **y = √2/2** and notice that we should have four solutions to our equation on this interval, one for every intersection of the line an the curve.

Now all that's left is to recognize that these solutions will have a repeating pattern. For each cycle of the sine curve, there will be a solution at **π/4** radians and one at **3π/4** – that much we recall by knowing what one cycle of **sin(x)** looks like. The other two solutions are just **π/4** and **3π/4** rad away from the start of the second cycle, at **2π**, so they're **2π + π/4** and **2π + 3π/4**.

So this equation has four solutions on the interval **[0, 4π]**. They are:

Moving the 1 to the right gives us

And taking the root of both sides yields:

The first thing you might want to do from here is look for a double-angle identity, like

That would be fine, but if you recall that **cos(2x)** is just a transformed version of **cos(x)** with two complete cycles of the function between 0 and 2**π**. That means, for our purposes, that there is a complete cosine cycle between 0 and **π**, and because **cos(x)** is an even function [meaning **f(-x) = (x)**], it has mirror symmetry across the y-axis. Here's what the graph looks like:

Our first solution is **x = 0**; then it's just a matter of noting that there's a solution to this equation every **π/4** radians, so our solutions on the interval [-**π**, **π**] are

**xaktly.com** by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.