Goniometric equations

Goniometric equations are equations that contain some goniometric function, that is, a sine, cosine, tangent, or cotangent.

The basic goniometric equation

Consider the equation sin x = 0. For which x will this equation be valid? When is the sine equal to zero? Let's start by looking at the graph:

We find that sin x equals zero quite often. Specifically, for the values of −π, 0, π, 2π and many others. Since sine is a periodic function, there will be regular intervals between solutions of the equation - in this case intervals of length π. Therefore, the set of all solutions of this equation can be written as a set, let's denote it R, which we define as follows:

$$ R = {K\cdot\pi | K \in \mathbb{Z}} $$

It is thus all K-folds of Pi, where K is an integer. Thus we have described the set of all solutions of the equation sin x = 0.

Recall that the sine and cosine have a range of values on the interval <−1, 1>, so the equation sin x = 2 has no solution, since we are unable to find one x, for which the expression sin x would have a value greater than one. Of course, the equation 3 · sin x = 2 already has a solution, because we can divide the whole expression by three and the number on the right hand side will already be less than one. We would be looking for the roots of the equation sin x = 2/3.

Calculating a simple example

We solve simple goniometric equations by reading the value from a graph or the unit circle, for example, and finding the period to add to the result. For example, solve the equation

$$ \sin x = \frac12 $$

We have the graph of the sine function above, so we look to see when the curve has the value $\frac12$. We find that this is the case for 1/6π and 5/6π. I strongly recommend learning the basic table values of goniometric functions, without them even a peek at the graph won't help you.

Now we will need to determine the period to know how to construct the set of all solutions. The sine function has a period of 2π - if you look at the graph, you will see that every time after 2π the function starts to repeat.

So if there is one solution to the equation 1/6π, then any number 1/6π + 2Kπ, where K is an integer, must be a solution to the equation, because at that point the sine function has the same functional value. The 2K just indicates that we only want even integers. If you put any integer after K and multiply it by two, you get an even number. Exactly the same for the other root, 5/6π. At the points 5/6π, 5/6π + 14π, and 5/6π − 82π, the function sin x has the same functional value, so all 5/6π + 2Kπ will solve the equation.

Write the resulting set R, which will be the union of the two previous results:

$$ R = \left\{\frac16\pi+2K\pi|K\in\mathbb{Z}\right\} \cup \left\{\frac56\pi+2K\pi|K\in\mathbb{Z}\right\} $$

Graphically, we could represent the solution as the intersection of the graphs of the function sin x (red graph) and the function $y = \frac12$ (green graph).

For example, if we put K = 0 after K, then we would get the solution:

$$\begin{eqnarray} x_1 &=& \frac16\pi+2K\pi = \frac16\pi+2\cdot0\cdot\pi=\frac16\pi\\ x_2 &=& \frac56\pi+2K\pi = \frac56\pi+2\cdot0\cdot\pi=\frac56\pi \end{eqnarray}$$

That is, x₁ = 1/6π and x₂ = 5/6π. In the figure, the corresponding points are B and C, because their x-coordinates are just 1/6π and 5/6π, respectively. If we substituted K = 1, we would get the points D and E.

Substitute

If we have a more complex expression in a function, we can use substitution, or replacement. For example, if we want to calculate the result of the equation sin 2x = 1, we substitute (make a substitution) a = 2x and then we calculate the equation in the form sin a = 1 just as we showed in the previous chapter. Substitution is usually done by taking the arguments of a function, in this case 2x, and substituting some other unknown in their place. We can name it pretty much anything, here we chose the name a. We could easily substitute q = 2x or pomeranč = 2x and then write sin pomeranč = 1.

The substitution thus simplifies the expression we are currently computing. We'll deal with the fact that the expression a is actually equal to 2x later.

So now we solve the equation sin a = 1, with a being the unknown. So we are looking for when the sine is equal to one. It turns out that the sine equals one when π/2 + 2Kπ. The roots of sin a = 1 are thus the values of the set

$$ R = \left\{\frac{\pi}{2}+2K\pi|K\in\mathbb{Z}\right\}. $$

Finally, we still need to remember our substitution - we need to undo it. We know that the solution to sin a = 1 is the elements of the previous set R. But we want to know the solution to another equation, sin 2x = 1. Yet we know that a = 2x.

We try to take one particular solution from the set of solutions R, for example for K = 1 we get a particular solution a = π/2 + 2π. At this point the equation sin a = 1 has a solution. Since a = 2x, it is also true that 2x = π/2 + 2π (we just wrote 2x instead of a ).

But we are not interested in 2x, we are interested in the value of x, so we divide the whole equation by two. This gives us the equation x = π/4+π. We have thus obtained one particular solution to the equation sin 2x = 1.

If we want to get all the solutions, we have to do this modification with all the elements of the set R. We construct the equation as follows:

$$ 2x = \frac{\pi}{2} + 2K\pi $$

and isolate x, which means we divide the whole equation by two:

$$ x = \frac{\pi}{4} + K\pi $$

This is the resulting solution. Let's write it down as a new set R':

$$ R' = \left\{\frac{\pi}{4} + K\pi|K\in\mathbb{Z}\right\} $$

Again, we can check the correctness of the solution graphically. The red is the graph of the function sin 2x and the green is the function y = 1.

If we substitute K = 0 after K, we get x = π/4, which represents the point C. If we substitute K = 2, we get x = π/4 + 2π, which represents the point E.

The formula

Adding with goniometric equations often uses a variety of formulas that you should be familiar with. Some are basic, others are more complicated. The goniometric formulas have moved to their own page, so take a look there.

Examples

1. Solve the equation sin (3x−π/2) = 0.

   Use the simple substitution a = 3x−π/2 to solve the equation sin a = 0. We already know that the sine is equal to zero for a = Kπ. So we add back the substitution and calculate the equation:

   $$ 3x-\frac{\pi}{2} = K\pi $$

   First, we convert π/2 to the left-hand side, i.e., add π/2 to the equation :

   $$ 3x = K\pi + \frac{\pi}{2} $$

   Now divide the equation by three:

   $$ x = \frac{K\pi}{3} + \frac{\pi}{6} $$

   The result is thus the set R₁, defined as follows:

   $$ R_1 = \left\{\frac{K\pi}{3} + \frac{\pi}{6}|K\in\mathbb{Z}\right\} $$

   Figure again:

   

   For K = 1 we get x = π/3 + π/6 = π/2, i.e. the point E.

2. Solve the equation cos²x − sin x = 1.

   First we use the formula and decompose cos² x into 1 − sin² x. Substitute back into the equation and we get:

   $$\begin{eqnarray} 1-\sin^2x-\sin x &=& 1\\ \sin^2x+\sin x &=& 0 \end{eqnarray}$$

   Now we plot sin x:

   $$ \sin x (\sin(x) + 1) = 0 $$

   Since we have got the left side of the equation in product form, and we have zero on the right side, we will solve the equations sin x = 0 and sin(x) + 1 = 0 separately , since the entire left side will be zero if at least one factor is zero.

   We have already solved the equation sin x = 0 several times, the solution is x of the form Kπ, where K is an integer.

   We convert the second equation, sin(x) + 1 = 0 to the equation sin(x) = −1. The sine is equal to minus one for the values of 3/2π + 2Kπ.

   All elements of the resulting set R₂ are then obtained by union.

   $$ R_2 = \left\{\frac32\pi + 2K\pi|K\in\mathbb{Z}\right\} \cup \left\{K\pi|K\in\mathbb{Z}\right\} $$

   Figure:

   

   We see that the points A, C, D, and F have x-coordinates multiple of π, so these points are described by the right set in the previous unification. And if we substitute the value K = 0 for K in the left set in the union, we get x = 3π/2, which corresponds to the point E, and if we substitute K = −1, we get x = 3π/2 − 2π = π/2, which corresponds to the point B.

3. Calculate the equation cos² x − sin²x + cos x = 0.

   In the first step, we apply the formula sin²x = 1−cos²x and decompose the sine squared:

   $$\begin{eqnarray} \cos^2x-(1-\cos^2x)+\cos x = 0\\ \cos^2x -1+\cos^2x+\cos x = 0\\ \end{eqnarray}$$

   In the next step, we add it together:

   $$ 2\cos^2x+\cos x-1=0 $$

   Finally, we make the substitution a = cos x.

   $$ 2a^2+a-1=0 $$

   We already solve this as a simple quadratic equation. Results:

   $$ a_1 = -1, a_2=\frac12 $$

   Finally, we undo the substitution and solve the pair of equations cos x = −1 and $\cos x = \frac12$ and unify the resulting sets. This is already simple and we have already solved it above.

4. Solve the equation sin²x−cos²x = 1.

   We first apply the classical formula and express the cosine in terms of the sine. We get:

   $$ 2\sin^2x=2 $$

   Divide by two and subtract one:

   $$ \sin^2x-1=0 $$

   Now we apply the formula a² − b² = (a − b)(a + b):

   $$ (\sin(x) - 1)\cdot(\sin(x) + 1) = 0 $$

   The product of two terms is zero if at least one of the terms is zero. We already know when sin x = 1 and sin x = −1. The solution to the goniometric equation is:

   $$ R_4 = \left\{\pi/2+2K\pi|K\in\mathbb{Z}\right\} \cup \left\{3\pi/2+2K\pi|K\in\mathbb{Z}\right\} = \left\{\pi/2+K\pi|K\in\mathbb{Z}\right\} $$

   Figure: