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# Difference between revisions of "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.

Third example:

Again, solve for $x$ in the domain $[0,2\pi]$.

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

Worry about the fraction later. First, move $cos^2(x)$ to the left.

$sin^2(x) + cos^2(x) + tan^2(x) = \frac{1}{sin^2(x) + cos^2(x)}$

Get rid of the fraction. No multiplying needed!

$sin^2(x) + cos^2(x) + tan^2(x) = \frac{1}{1}$

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

Now we can use an identity. I'm not talking about $sin^2(x) + cos^2(x) = 1$ (though that is a possibility) I'm referring to the identity $sin^2(x) + cos^2(x) + tan^2(x) = sec^2(x)$.

$sec^2(x) = 1$

Now take the $-1$th power of each side of the equation:

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

Take the square root:

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

For $1$, $x$ can equal $0$ or $\2pi$ (Error compiling LaTeX. ! Undefined control sequence.). For $-1$, $x$ can equal $\pi$. Those our the three solutions.

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