Periodic functions → many solutions

Often we are faced with solving equations involving trigonometric equations, like $\sin(x) = 1/2$. Because of the periodic nature of the trigonometric functions – they repeat themselves infinitely many times – the variable x can take on an infinite number of solutions.

While it is possible to communicate to a reader the infinite set of all solutions to such equations, , we usually just restrict the domain to only a small region so that the number of solutions is finite. For example, we might need the solutions to $sin(x) = 1/2$ on the interval $[0, \, \pi]$ or the interval $(-\pi/2, \, \pi/2)$.

Tools you'll need to solve trig equations

In order to be effective solving trigonometric equations, there are some things you'll just have to know first. If you don't, now's the time to review them.

• You have to know the sines, cosines and tangents of the common angles (0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, 2π) in radians.


• It's crucial that you are able to make 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.


• You should 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 pieces.


• You'll need to know your analytic trigonometry, mainly the Pythagorean identities and a few others that can be very helpful. At least be able to quickly look up the double angle and trig. sum formulas.

Example 1

Solve the equation $\tan(x) = \sqrt{3}$ on the interval $[-2\pi, \, 2\pi]$.

Solution: In this problem, we recognize that $\sqrt{3}$ 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 $[-2\pi, \; 2\pi]$. On it, we'll plot our initial solution, $x = \sqrt{3}/3.$

Example 2

Solve the equation $2 \sin(x) = \sqrt{x}$ on the interval $[0, \, 4\pi]$.

Solution: We begin by looking at the equation. Nothing really changes that much in algebraic problems, and this this one we're trying to solve for $x$, so we might as well "peel away" the stuff that's easiest to remove.

$$2 \sin(x) = \sqrt{2}$$

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

$$\sin(x) = \frac{\sqrt{2}}{2}$$

Now hopefully that $\sqrt{2}/2$ 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

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

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

Example 3

Solve the equation $cos^2(2x) - 1 = 0$ on the interval $[-\pi, \, \pi]$.

Solution: We'll again use a few basic algebra moves to first simplify this function.

$$\cos^2(2x) - 1 = 0$$

Moving the 1 to the right gives us

$$\cos^2(2x) = 1$$

And taking the root of both sides yields:

$$\cos(2x) = ± 1$$

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

$$\cos(2x) = \cos^2(x) - \sin^2(x)$$

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 $2 \pi$;.

Practice problems

Solve the following trigonometric equations analytically on the interval $[0, \, 2\pi]$

1. $$\sin(x) = 1/2$$

   Solution

   The 1/2 should bring to mind the 30-60-90 triangle, which shows us that the sine of 30˚ (π/6) is 1/2.

   

   The obvious solution is π/6, but if we sketch out sin(x) we see that there are three more solutions on this interval, labeled A, B and C

   

   Finding the other solutions is a matter of using the symmetry of the sine function. Solution B is π/6 units back from π, and so forth:

   

   The solutions are

   $$x=\frac{\pi}{6}, \; \frac{5\pi}{6}, \; \frac{7\pi}{6}, \; \frac{17\pi}{6}$$




2. $$\sqrt{3} \cdot \text{cot}(x) + 1 = 0$$

   Solution

   A good strategy for a problem like this is to isolate the trig. function, which gives.

   $$\text{cot}(x) = \frac{-1}{\sqrt{3}}$$

   Now the cotangent might be awskard to work with, so take the reciprocal of both sides to convert to tangent (you should know how to sketch several cycles of tan(x)):

   $$\tan(x) = -\sqrt{3}$$

   Now the $\sqrt{3}$ should remind you of the 30-60-90 triangle, for which the tanbent of π/6 is $\sqrt{3}$:

   

   Now we want $-\sqrt{3}$, but that's not to hard: just sketch out the tangent function and notice thhat it has some symmetry: tan(-π/6)=$-\sqrt{3}$. so that's our first, or "prototype" solution:

   

   All solutions are obtained by going to the zero of each tan(x) curve and "backing up"e; (subtracting) π/6 units. The only ones in the [0, 2π] interval are:

   $$x = \frac{5\pi}{6}, \; \frac{7\pi}{6}$$




3. $$2 \, \sin(x) + \sqrt{3} = 0$$

   Solution

   
   

   A good strategy for a problem like this is to isolate the trig. function, which gives

   $$\sin(x) = \frac{-\sqrt{3}}{2}$$

   Now the $\sqrt{3}{2}$ should remind you of a 30-60-90 triangle, for which the sine of π/3 is $\sqrt{3}/2$.

   

   Now we want $-\sqrt{3}{2}$, but that's not too hard — just sketch out the sine function and notice that it has some symmetry: sin(-π/3) = $-\sqrt{3}{2}$. That solution is out of our domain, but notice that it's π/3 units from the zero of sin(x):

   

   so there are two solutions on the interval [0, 2π]. One is π/3 to the right of zero, the other π/3 to the left of 2π

   

   $$x=\frac{4\pi}{3}, \; \frac{5\pi}{3}$$




4. $$\text{sec}^2(x) - 1 = 0$$

   Solution

   
   

   First isolate the trig. function on the left:

   $$\text{sec}^2(x)=1$$

   Take the root of both sides (dont' forget the $\pm$).

   $$\text{sec}(x)=\pm 1$$

   Now convert to one of the big three trig. functions (sin, cos, tan) to make it easier to visualize the solutions:

   $$ \begin{align} \frac{1}{\cos(x)} &=\pm 1 \\[5pt] \cos(x) &= \pm 1 \end{align}$$

   Now sketch the cosine function and notice that there are three places where cos(x) = ±1 on the interval [0, 2π]

   

   So the solutions are

   $$x=0, \; \pi, \; 2\pi$$