Trigonometric Equations and Inequalities
Why this matters
Imagine you're managing a new waterfront park in Seattle. Your biggest attraction is a giant Ferris wheel. For a special July 4th fireworks display, you want the ride to stop when certain cabins are exactly 80 feet high, offering the best view. The height of a cabin, h, can be modeled by a sine or cosine function based on the angle of rotation, θ.
The question "At what angles θ is the height h exactly 80 feet?" is a trigonometric equation. It's not a one-time thing. The wheel keeps turning, so there will be multiple angles (and multiple times) when a cabin hits that perfect height.
Solving these equations is the key. It's how we find all the perfect moments, whether for a fireworks show, predicting daylight hours, or modeling sound waves. Today, we'll learn the systematic way to solve these repeating puzzles.
Concept overview
flowchart TD
A[Start: Trig Equation, e.g., 2cos(x) + 1 = 0] --> B{1. Isolate the trig function};
B --> C[cos(x) = -1/2];
C --> D{2. Find Principal Solution using inverse trig};
D --> E[x = arccos(-1/2) = 2π/3];
E --> F{3. Find other solutions in [0, 2π] using symmetry};
F --> G[For cosine, use 2π - x --> x = 2π - 2π/3 = 4π/3];
G --> H{4. Write General Solution (add period * k)};
H --> I[x = 2π/3 + 2πk <br> x = 4π/3 + 2πk];
I --> J{5. Apply domain restrictions, if any};
J --> K[Solutions in [0, 2π] are 2π/3, 4π/3];
Core explanation
You've spent a lot of time solving algebraic equations. If I gave you 2y - 1 = 0, you'd know exactly what to do: add 1 to both sides, divide by 2, and get y = 1/2.
Solving trigonometric equations starts the exact same way.
Let's take a look at a classic example:
2sin(x) - 1 = 0
Step 1: Isolate the Trigonometric Function
Just like you isolated y in the algebra problem, your first goal is to get the sin(x) part by itself.
- Add 1 to both sides:
2sin(x) = 1 - Divide by 2:
sin(x) = 1/2
Great! Now we have a simpler question: "For what angle x is the sine equal to one-half?"
Step 2: Find the First Solution (The "Calculator Answer")
This is where your knowledge of inverse trig functions comes in. To find x, we can take the arcsin of both sides:
x = arcsin(1/2)
If you plug this into your calculator (in radian mode!), you'll get:
x = π/6
This is called the principal value. It's a perfectly correct answer, but it's only one piece of the puzzle.
Step 3: Find All Solutions in One Full Period
Think about the unit circle. The sine of an angle is its y-coordinate. We're looking for all the places where the y-coordinate is 1/2.
(Note: A real image would be here)
You'll see one angle in Quadrant I, which is π/6 (our calculator answer). But there's another angle over in Quadrant II that also has a positive y-coordinate of 1/2.
How do we find it? Because of the symmetry of the sine wave and the unit circle, this second angle is π - x.
So, the second solution is:
x = π - π/6 = 5π/6
Now we have the two solutions within the first full rotation, or the interval [0, 2π]: x = π/6 and x = 5π/6.
Step 4: Write the General Solution
The sine function repeats its values every 2π radians. So, if π/6 is a solution, so is π/6 + 2π, π/6 + 4π, π/6 - 2π, and so on.
We capture this infinite family of solutions by adding 2πk, where k is any integer (..., -2, -1, 0, 1, 2, ...).
So, the complete, general solution is actually two sets of equations:
x = π/6 + 2πkx = 5π/6 + 2πk
This covers every single angle, in all of eternity, for which sin(x) is 1/2.
What About Context and Limited Domains?
Most of the time, the AP exam or a real-world problem won't ask for all infinite solutions. They'll give you a specific interval, like "Solve 2sin(x) - 1 = 0 on the interval [0, 2π]."
In that case, you'd perform steps 1-3 and find that the only answers that fall within that specific window are π/6 and 5π/6.
If the interval was, say, [-π, π], you'd use your general solution formulas and plug in different integer values for k to see which results land in the interval.
- For
k=0:π/6and5π/6(both are in[-π, π]) - For
k=1:π/6 + 2π(too big) and5π/6 + 2π(too big) - For
k=-1:π/6 - 2π = -11π/6(too small) and5π/6 - 2π = -7π/6(too small) So for the interval[-π, π], the solutions areπ/6and5π/6.
A Quick Note on Inequalities
What if the problem was 2sin(x) - 1 > 0?
First, you'd isolate the trig function to get sin(x) > 1/2.
The trick is to first solve the equality sin(x) = 1/2 to find the boundary points, which we already know are π/6 and 5π/6 in the first period.
Then, look at the graph of y = sin(x). Where is the curve above the line y = 1/2? It's between our two boundary points. So, the solution on [0, 2π] would be the interval (π/6, 5π/6).
Worked examples
Solving a Cosine Equation
Problem: Find all solutions to 2cos(x) + √3 = 0 on the interval [0, 2π].
Step 1: Isolate the cosine function.
Subtract √3 from both sides:
2cos(x) = -√3
Divide by 2:
cos(x) = -√3 / 2
Step 2: Find the principal value.
Using the inverse cosine function, x = arccos(-√3 / 2). Your calculator will give you the principal value, which is in the range [0, π].
x = 5π/6
Step 3: Find other solutions in the interval [0, 2π].
Cosine represents the x-coordinate on the unit circle. We're looking for where the x-coordinate is -√3 / 2. We found 5π/6 in Quadrant II. Where else is the x-coordinate negative? In Quadrant III.
For cosine, the symmetry rule is 2π - x for the other angle. So, our second solution is:
x = 2π - 5π/6 = 12π/6 - 5π/6 = 7π/6
Step 4: Check if the solutions are in the domain.
The problem asks for solutions on [0, 2π]. Our two solutions, 5π/6 and 7π/6, are both within this interval.
Final Answer: The solutions are x = 5π/6 and x = 7π/6.
Solving an Equation with Factoring
Problem: Find all solutions to sin(x)cos(x) - sin(x) = 0 on the interval [0, 2π].
Step 1: This is where students make a huge mistake. They might try to divide both sides by sin(x). Do not do this! If you divide by a variable expression, you might be dividing by zero and you will lose solutions.
The correct approach is to factor. Move all terms to one side (already done) and factor out the common term, which is sin(x).
sin(x) * (cos(x) - 1) = 0
Step 2: Apply the Zero Product Property.
Just like in algebra, if A * B = 0, then either A = 0 or B = 0. So we have two simpler equations to solve:
sin(x) = 0cos(x) - 1 = 0, which simplifies tocos(x) = 1
Step 3: Solve each simple equation on the interval [0, 2π].
- For
sin(x) = 0: On the unit circle, where is the y-coordinate zero? Atx = 0andx = π. (Note:2πis the end of the interval, but it's the same position as0, so we typically list0.) - For
cos(x) = 1: On the unit circle, where is the x-coordinate one? Only atx = 0.
Step 4: Combine the unique solutions.
Our collection of solutions is {0, π, 0}. We only list each unique solution once.
Final Answer: The solutions are x = 0 and x = π.
Try it yourself
Ready to try a couple on your own? Don't just go for the answer; walk through the steps we discussed.
- 1ProblemFind all solutions for
4cos²(θ) - 3 = 0on the interval[0, 2π]. - 2ProblemFind the general solution for
tan(x) + 1 = 0.
Good luck! You've got this.
Practice — 8 questions
In simple terms, this is about finding the angles that make a trigonometric statement true, like solving a puzzle with repeating waves to find where they hit a certain height.
- 3.10.A: Solve equations and inequalities involving trigonometric functions.
- 3.10.A.1
- Inverse trigonometric functions are useful in solving equations and inequalities involving trigonometric functions, but solutions may need to be modified due to domain restrictions.
- 3.10.A.2
- Because trigonometric functions are periodic, there are often infinitely many solutions to trigonometric equations.
- 3.10.A.3
- In trigonometric equations and inequalities arising from a contextual scenario, there is often a domain restriction that can be implied from the context, which limits the number of solutions.
flowchart TD
A[Start: Trig Equation, e.g., 2cos(x) + 1 = 0] --> B{1. Isolate the trig function};
B --> C[cos(x) = -1/2];
C --> D{2. Find Principal Solution using inverse trig};
D --> E[x = arccos(-1/2) = 2π/3];
E --> F{3. Find other solutions in [0, 2π] using symmetry};
F --> G[For cosine, use 2π - x --> x = 2π - 2π/3 = 4π/3];
G --> H{4. Write General Solution (add period * k)};
H --> I[x = 2π/3 + 2πk <br> x = 4π/3 + 2πk];
I --> J{5. Apply domain restrictions, if any};
J --> K[Solutions in [0, 2π] are 2π/3, 4π/3];
Read what Saavi narrates
Hello everyone, it's Saavi from Shrutam.
Imagine you're managing a new waterfront park in Seattle. Your biggest attraction is a giant Ferris wheel. For a special July 4th fireworks display, you want the ride to stop when certain cabins are exactly 80 feet high, offering the best view. The height of a cabin can be modeled by a sine or cosine function. So, the question "At what angles is the height exactly 80 feet?" is a trigonometric equation. And since the wheel keeps turning, there will be multiple angles... multiple moments... that are correct.
That's what we're doing today: solving those equations. We're working backward to find the angle 'x' that makes a statement like sine of x equals one-half true. But here's the twist... because these functions are waves, they repeat. So there's usually more than one answer.
Let's walk through an example together. Let's find all solutions to the equation two times cosine of x, plus the square root of three, equals zero... and we'll do it on the interval from zero to two pi.
First, we need to get cosine of x by itself. So, we'll subtract the square root of three from both sides, and then divide by two. That gives us... cosine of x equals negative square root of three, over two.
Okay, now what angle has a cosine of negative root three over two? We use the inverse cosine. Your calculator will tell you the principal value is five pi over six.
And here is the number one mistake I see students make every single year. They find that one answer, five pi over six, and they stop. But remember the unit circle! Cosine is the x-coordinate. It's negative in Quadrant Two... but it's also negative in Quadrant Three. There's another answer! For cosine, we find it by doing two pi minus our first answer. So, two pi minus five pi over six gives us seven pi over six.
So, the two solutions on that interval are five pi over six and seven pi over six.
See? It's a process. Isolate, find the first answer, then use symmetry to find the second. You can do this. Keep practicing, and it will become second nature.
Trigonometric functions are periodic. For every one solution your calculator gives, there's often another one (or more) within a single period, plus infinite solutions in general.
After finding `x = arcsin(y)`, always check the unit circle or use symmetry rules (`π - x` for sine, `2π - x` for cosine) to find the second solution in the `[0, 2π]` interval.
Sine and cosine have a period of `2π`, but tangent and cotangent have a period of `π`. Adding `+ 2πk` to a tangent solution will miss half of the answers.
Memorize the periods. For sine/cosine/secant/cosecant, add `+ 2πk`. For tangent/cotangent, add `+ πk`.
If you solve `sin(x)cos(x) = sin(x)` by dividing by `sin(x)`, you assume `sin(x)` is not zero. This makes you lose all the solutions where `sin(x)` *is* zero.
Bring all terms to one side of the equation and factor. `sin(x)cos(x) - sin(x) = 0` becomes `sin(x)(cos(x) - 1) = 0`. Then set each factor to zero.
`arcsin(0.5)` is `30°` in degree mode but `π/6` (which is about 0.524) in radian mode. Using the wrong mode will give you a completely incorrect numerical answer.
Before any test or quiz, get in the habit of checking your calculator's mode. AP Precalculus and Calculus almost always use radians.
Using `π - x` for cosine or `2π - x` for sine will lead to an incorrect second solution.
Think of the unit circle. Sine is the y-coordinate, which is symmetric across the y-axis (so `x` and `π-x` have the same sine). Cosine is the x-coordinate, symmetric across the x-axis (so `x` and `-x` or `2π-x` have the same cosine).