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 2π. 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
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