# Trigonometric Equations

It is probably good to know how to solve trigonometric equations, which often involved brute force and the use of trigonometric identities.

Here is an example:

Solve for $x$ for all answers in the domain $[0, 2\pi]$.

$sin^2(x) - 1 = cot^2(x)$

Note the quadratics, and more implortantly, the $-1$. The 1 is one end of a very well-known trigonometric identity, and substituting it with the other end makes this equation much easier to solve.

$sin^2(x) - (sin^2(x) + cos^2(x)) = cot^2(x)$

The obvious next step is distributing:

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

Next, get rid of the $sin^2(x)$:

$-cos^2(x) = cot^2(x)$

What next? Recall how $cot^2(x)$ can be rewritten using an identity. Look below:

$-cos^2(x) = \frac{cos^2(x)}{sin^2(x)}$

The $cos^2(x)$s on both sides can be divided away, and the result is

$-1 = \frac{1}{sin^2(x)}$

To get rid of the fraction, both sides of the equation can be taken to the power of $-1$:

$\frac{1}{-1} = sin^2(x)$

$-1 = sin^2(x)$

Next, take the square root of both sides of the equation:

$\sqrt{-1} = \sqrt{sin^2(x)}$

$i = sin(x)$

We have a problem! The domain for values of $x$ is $[0, 2\pi]$. However, no real value of $x$ can become imaginary when put into the function $f(x) = sin(x)$. So, there are no solutions.

Here is another problem:

$sin^2(x) + sin(x) + sin(2x) = 0$

Again, we are solving for $x$ in the domain $[0, 2\pi]$

Hey, there are only sines here! Too bad it's not being kept that way.

The first step is to use the identity $sin(2x) = 2sin(x)cos(x)$, as this gets rid of the $2x$ inside the trigonometric function.

$sin^2(x) + sin(x) + 2sin(x)cos(x) = 0$

Every term on the left side now has a $sin(x)$. We can now factor that out of the left side of the equation, resulting in

$sin(x)(sin(x) + 1 + 2cos(x)) = 0$

We can work with both parts of the left side to find solutions. Let's do the shorter one first.

$sin(x) = 0$

When solving trigonometric equations, it probably doesn't get easier than this. Using the unit circle or a graph of sin(x), we can see that the solutions in the given domain are $0$, $\pi$, and $2\pi$.

Now we can work with the other part, which is

$sin(x) + 1 + 2cos(x) = 0$

Let's get the trigonometric functions on one side and the constants on the other. That can be done be subtracting both sides of the equation by $1$:

$sin(x) + 2cos(x) = -1$

Now, consider this. What is the square of $-1$? $1$! What equals $1$? $sin^2(x) + cos^2(x)$! By squaring both the left and right sides of the equation, we might be able to put that identity to use. Let's do that!

$(sin(x) + 2cos(x))^2 = (-1)^2$

$(sin(x) + 2cos(x))(sin(x) + 2cos(x)) = 1$

$sin^2(x) + 2sin(x)cos(x) + 2sin(x)cos(x) + 4cos^2(x) = 1$

Combining some terms:

$sin^2(x) + 4sin(x)cos(x) + 4cos^2(x) = 1$

Using that identity, $1 = sin^2(x) + cos^2(x)$:

$sin^2(x) + 4sin(x)cos(x) + 4cos^2(x) = sin^2(x) + cos^2(x)$

We can move the rewritten $1$ to the left side and eliminate the $sin^2(x)$ terms completely!

$4sin(x)cos(x) + 3cos^2(x) = 0$

We can factor out a $cos(x)$ that is common to all of the terms on the left side:

$cos(x)(4sin(x) + 3cos(x)) = 0$

We can just work with the left part again, $cos(x)$, equal to $0$.

$cos(x) = 0$

With the help of the Unit Circle, we see that the solutions for x that are found with this in our domain are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.

Now we can work with the right part.

$4sin(x) + 3cos(x) = 0$

To square the equation for some eliminating without a middle term, first, let's move the $3cos(x)$ to the other side.

$4sin(x) = -3cos(x)$

Now we can square the equation:

$(4sin(x))^2 = (-3cos(x)^2$

$16sin^2(x) = 9cos^2(x)$

We note that both $16$ and $9$ are perfect squares, and we could move the $9cos^2(x)$ back and do a Difference of Squares factorization and solve for $x$ that way. However, there is an easier way that lets us work with only one trigonometric function instead of two. Let's move $9cos^2(x)$ back first:

$16sin^2(x) - 9cos^2(x) = 0$

Now we can use the $sin^2(x) + cos^2(x) = 1$ identity again. How? It is not by trying to factor the $sin^2(x) + cos^2(x)$ out: that would probably just leave a more complicated expression on the left side than we have now. Instead, try subtracting both sides of this identity by $sin^2(x)$.

We have $cos^2(x) = 1 - sin^2(x)$

We just proved this well-known identity with ease. Since we proved it we can use it, like this:

$16sin^2(x) - 9(1 - sin^2(x)) = 0$

Distributing:

$16sin^2(x) - 9 + 9sin^2(x) = 0$

We can combine the $sin^2(x)$ terms and move the $-9$ to the other sides of the equation.

$25sin^2(x) = 9$

Now we can divide both sides of the equation by $25$:

$sin^2(x) = \frac{9}{25}$

Note the numerator and denominator of the fraction in this equation. The fraction can be rewritten in a way that makes it clear what to do next:

$sin^2(x) = (\frac{3}{4})^2$

Take the square root of both sides of the equation and get

$sin(x) = \pm \frac{3}{4}$

We can take the arcsine of both sides of the equation (with a calculator!) of both the positive and negative $\frac{3}{4}$

Arcsin of $\frac{3}{4}$: approximately $.848062079$

Arcsin of $-\frac{3}{4}$: approximately $-.848062079$

However, consider the identity $sin(x) = sin(\pi - x)$

This indicates two more solutions: $\pi$ minus our other solutions. Those give us

$3.989654733$

and

$2.293530575$

So, we have all nine of those solutions for our original equation.

You may come across far more difficult trigonometric equations than these, but hopefully from reading this, you are well-equipped for those challenges.