Before you start to learn about trigonometric equations, read and learn all about radicals and exponents. Without knowing everything about it, you’ll not know methods of solving trigonometric equations and inequalities.
Ok, let’s start.
Trigonometric equations are equations that come in form $ f(x) = a$, where a is a real number and $ f(x)$ some trigonometric function.
Every trigonometric equation will have infinitely many solutions, but since trigonometric functions are periodic, we can write them down as a set of solutions in a relatively pretty way.
1. Trigonometric equation $ cos x = a$, $ \mid a \mid \le 1$
The set of solutions, B, will be: $ B = (\pm x + 2k\pi, k \in \mathbb{Z})$, where x is one solution of that equation.
Example 1.: Find the $ x$ in $ cos(x) = 1$
$ Cos(x) = 1$ tells us that the value for cosine. This is true only for $ 0 +2 k\pi$. You don’t have to remember anything, just draw a unit circle and conclude from it.
Example 2.: Find $ x$ in $ cos(x) = \frac{1}{2}$
From this you can conclude that there are two angles whose sine value is equal to $\frac{1}{2}$. $ X$ and $ -X$.
How would you find it?
If $ cos(x) = \frac{1}{2}$ that means that $ x = Arc Cos(\frac{1}{2})$ and if we remember our basic trigonometric value table, we can see that x is equal to $\frac{\pi}{3}$. Now we found only one solution, the final set of solutions will be:
$ B = ( \pm \frac{\pi}{3} + 2k\pi, k \in \mathbb{Z} )$
2. Trigonometric equation $ sin(x) = a$, $ \mid a \mid \le 1$
The set of solutions, B, will be: $ B = ( x + 2k\pi, k \in \mathbb{Z})$ $\bigcup$ $(\pi – x + 2k\pi, k \in \mathbb{Z})$, where $x$ is one solution of that equation.
Why this form of solution. If you take a look at the unit circle and examine for which angles are the sine values equal you’ll notice that you also have two angles:
$ arc sin(\frac{\sqrt{2}}{2}) = x$, $ x = \frac{\pi}{4}$, this is one solution we got. And to write it as a whole set of solutions:
$ B = (\frac{\pi}{4} + 2k\pi, k \in \mathbb{Z})$ U $(\frac{3 \pi}{4} + 2k\pi, k \in \mathbb{Z})$
3. Trigonometric equations $ tan x = a$, $ a \epsilon R$
The set of solutions, B will be $ B = (x + k\pi, k \in \mathbb{Z})$
Example 1.: solve $ tan x = 1 \rightarrow x = \frac{\pi}{4} \rightarrow B = (\frac{\pi}{4} + k\pi, k \in \mathbb{Z})$
4. Trigonometric equations $ cot x = a$, $ a \in \mathbb{R}$
The set of solutions, B will be $ B = (x + k\pi , k \in \mathbb{Z} )$
Each trigonometric equations have a one method that is the best for finding a solution. To know what method to use, you’ll need to learn about each method and then decide which one you will use for particular equation.
Method of solving trigonometric equations using substitution
When the argument of any trigonometric function is more than just the unknown we can substitute that whole expression. When we solve the values for our substitution we go back to the starting substitution and find our main angle.
First something simple:
$ sin(2x) = 1$. Here we’re not trying to mainly find the angle whose sine value is equal to 1, but to find x.
The substitution must help us by simplifying this task into something we know how to solve, for example $ sin t = 1$. Which means that we’ll use substitution $ t = 2x$.
Now we got a whole new equation:
$ Sin t = 1 \rightarrow t = Arcsin(1)$. This is a tabular value, and from here $ t = \frac{\pi}{2}$. If you are unsure about your knowledge about tabular values solution, you can check it on your calculators. To get to the arcus sinus you have to press SHIFT and then sin, and enter the value. The value must be equal to the angle you got. If you want to turn it into radians you’ll have to divide with π. When you calculate $ arcsin(1)$ you’ll get 1.5707 and when you divide that with $ \pi$ you’ll get $\frac{\pi}{2}$, just make sure your calculator is set to radians. If it’s easier to you to first calculate in degrees, you can, but later it’s easier to calculate with radians (we learned transforming degrees into radians and vise versa in line segments and angles).
We’ll mark the solutions for the equations $ sin t = 1$ with $ B_t$, and for $ sin(2x) = 1$ with $ B_x$.
Again the solution is not only this one angle it has to be a set.
$ B_t = ( \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z}) \bigcup (\pi – \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z})$$ = ( \frac{\pi}{2}+ 2k\pi, k \in \mathbb{Z})$ $\bigcup$ $(\frac{\pi}{2} + 2k\pi, k \in \mathbb{Z})$
These two sets are the same so we can write only $ B_t = ( \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z} )$.
We’re not done yet. Don’t forget what you are looking for. We need to find x and the final set of solutions.
$ 2 B_x = B_t$, now every addend in $ B_t$ must be divided by 2.
$ B_x = ( \frac{\pi}{4} + \frac{k\pi}{2}, k \in \mathbb{Z})$ which is our final solution.
Transformation of trigonometric equations to polynomial
Form of $ P(f(x))$ where P is a polynomial, and $ f(x)$ is a trigonometric function.
These kind of problems are solved using substitution and turning our equation into a polynomial we know. Be careful, these kinds of substitutions can only be used if you have only one type of trigonometric function in your equation, sine, cosine, tangent or cotangent.
For example $ 2 sin^2(x) – 3 sin(x) + 1 = 0$
As you can notice this looks a lot like a quadratic equation, but we have a function as a variable.
For this we’ll use substitution $ t = sin(x)$ and get:
$ 2t^2 – 3t + 1 = 0$ and we got equation we know how to solve.
We got two solutions for t and now we get back to the substitution:
For $ x_1 = 1$:
$ t = sin(x) \rightarrow sin(x) = 1$
$ B_1 = (\frac{\pi}{2} + 2k\pi, k \in \mathbb{Z} )$.
And for $ x_2 = \frac{1}{2}$
$ t = sin(x) -> sin(x) = \frac{1}{2}$
$ B_2 = (\frac{\pi}{6}+ 2k\pi) U (\frac{5 \pi}{6}+ 2k\pi)$
The final solution is $ B_1 U B_2$:
$( \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z}) \bigcup (\frac{\pi}{6}+ 2k\pi) U (\frac{5 \pi}{6}+ 2k\pi)$.
Transformation of trigonometric equations to sin(x) + b cos(x) = c (the universal substitution)
Till now, in every equation was only one trigonometric function, and that was simple, we would use simple substitution and got our solution. Now we can’t use that substitution because x is bound with different functions.
We know that sine and cosine can be connected by one function- tangent tan $ x = \frac{sin x}{cos x}$, and we can use that to get only one function for our substitution.
The universal substitution
$ x \in \mathbb{R} / (\pi + 2 k\pi)$ -> substitution $ t = tan(\frac{x}{2})$
Of course, the denominator cannot be zero, and because of that we have a condition that $ x \epsilon R / (\pi + 2 k\pi) x$
Example:
Solve the equation $ 3 sin(x) + 4 cos(x) = 1$
We use the universal substitution:
We can notice that this polynomial is the square of difference $ (3t – 1)^2 = 0 \rightarrow t = \frac{1}{3}$
And to get back to our substitution:
$ B = (0.105\pi +2k\pi : k \in \mathbb{Z})$
If the solution isn’t so “pretty” you’ll get two values for t, put both of them into the substitution, and unite the solutions.
Method of solving systems of trigonometric equations
The system of trigonometric equations is any system of equation in which at least one is trigonometric.
Solving these equations usually rely on trigonometric identities or simple adjustment.
Example: solve the system $ sin(x)sin(y) = 1$, $cos(y)cos(x) = 0$
If we add these two equations we get $ sin(x)sin(y) + cos(y)cos(x) = 1$
And this looks familiar. Here we can use our addition formula $ sinx siny + cosy cosy = cos (x – y)$
Which leads us to $ cos(x – y) = 1 \rightarrow x – y = arc cos1 \rightarrow x – y = 0 + 2k\pi$
If we subtract these two equations we get $ sinx \cdot sin(y) – cos(y) \cdot cos(y) = 1$, but this is not any trigonometry identity. We know that $ cos(x + y) = cos(x)cos(y) – sin(x)sin(y)$, to get to that we’ll multiply that equation with – 1 and get that $ cos(x + y) = -1 \rightarrow x + y = arccos(-1) \rightarrow x + y = \pi + 2n\pi$.
We got another system: $ x + y = \pi + 2n\pi$, $ x – y = 2k\pi$
if we add these two together: $ 2x = \pi + 2n\pi + 2k\pi \rightarrow x = \frac{\pi}{2} + (n + k)\pi \rightarrow \frac{\pi}{2} + (n + k)\pi + y = \pi + 2n\pi$
$ y = \frac{\pi}{2} + 2n\pi – n\pi – k\pi \rightarrow y = \frac{\pi}{2} + (n – k)\pi$
This is a very simple example of a system of trigonometric equation, but every other system is solved on an analog way. The only complication there can be is that for every equation you have you have 2 values (positive and negative) which will give you four solutions for one variable.
For example if we got:
$ x – y = \frac{\pi}{6}$ and $ x + y = \frac{\pi}{2}$
$ x – y = -\frac{\pi}{6}$ and $ x + y = \frac{\pi}{2}$
$ x – y = \frac{\pi}{6}$ and $ x + y = -\frac{\pi}{2}$
$ x – y = -\frac{\pi}{6}$ and $ x + y = -\frac{\pi}{2}$, just solve them, and you’ll get four solutions for x, and four for y.
Trigonometric inequalities
Every trigonometric inequality is solved by first solving its trigonometric equation and concluding from the unit circle. Just like in any other inequality, we’ll have one or more intervals for a solution.
Example 1. Solve inequality $ cos(x) \le \frac{1}{2}$
First step is to solve equation $ cos(x) = \frac{1}{2}$ to get to the endpoints of our interval.
These are $ B = (\pm \frac{\pi}{3} + 2k\pi, k \in \mathbb{Z})$.
Draw a unit circle and mark angles $\frac{\pi}{3}$ and $ -\frac{\pi}{3} = \frac{5 \pi}{3}$.
You want to find all angles whose cosine value is less than or equal to $\frac{1}{2}$. That means that we are traveling over the unit circle to the left from point $\frac{\pi}{3}$ and go over to the $\frac{5 \pi}{3}$. And we take all those angles, because their cosine value is less than $\frac{1}{2}$. If we continue from $\frac{5 \pi}{3}$ to the $\frac{\pi}{3}$, their cosine value will be greater than $\frac{1}{2}$. Which means that our one solution will be $ [\frac{\pi}{3}, \frac{5 \pi}{3}]$.
But since this will also be true for other angles that are for whole circle distant from them.
This leads us to our final solution
We read this as union by all whole numbers k, of segments $ [\frac{\pi}{3} + 2k\pi, \frac{5 \pi}{3} + 2k\pi]$.
This only means that we are uniting all segments with these property.
Example 2:
Solve the inequality $ tan(x) \le 1$
First $ tan x = 1$: solutions in first lap are $\frac{\pi}{4} + k\pi$ and $\frac{\pi}{2} + k\pi$.
Now we’re looking for all angles whose tangent value is less than or equal to 1. These angles are from 0 to $\frac{\pi}{4}$ and from $\pi$ to $\frac{5 \pi}{4}$. This leads us to our final solution:
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