# Trigonometric Equations

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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 $[0, 2\pi}$ (Error compiling LaTeX. ! Extra }, or forgotten $.). $sin^2(x) - 1 = cot^2(x)$ 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$(Error compiling LaTeX. ! Missing$ inserted.)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.