The last lesson of Chapter 10. An inequality asks where one side exceeds the other, and the answer is a union of intervals rather than a list of points. Covers the sign diagram technique with its two kinds of landmark, choosing test values in awkward gaps, using the same machinery to find the domain of any function built from circular functions, the strategy of solving on one period and translating, and equations and inequalities in the inverse trigonometric functions, which behave in the opposite way to everything else in the chapter.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.7 Trigonometric Equations and Inequalities, pp. 866-874
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 866-874 — the pages these objectives are drawn from
Warm-up
You can solve any trigonometric equation. An inequality is solved by first solving the equation, and then asking one extra question.
Discussion prompt
You have found that 2 sine x minus 1 is zero at pi over six and five pi over six. Without any further calculation, what do those two points do to the interval from zero to two pi, and what would you check next?
Hint: Think about where the expression could possibly change sign.
Answer:
They cut the interval into three pieces. And because the expression is continuous, it cannot change sign anywhere except at a zero — so within each piece it keeps one sign throughout.
What to check next is therefore just one test value per piece, three evaluations in total. That is the entire sign-diagram method: the zeros do the hard work of locating the changes, and the test values only have to identify which side of zero each region is on.
Concept
Between two consecutive zeros, a continuous expression cannot cross zero, so it keeps a constant sign. That reduces an inequality over an entire interval to a handful of test evaluations.
The technique is the same one used for polynomial and rational inequalities. What is new is that a trigonometric expression has infinitely many landmarks, so an interval must be stated or a period chosen.
Figure (svg): A sign diagram for two sine x minus one on the interval from zero to two pi, showing the expression negative before pi over six, positive between pi over six and five pi over six, and negative afterwards
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 866-866
Section
Section 1
Concept
Gather everything to one side, find the zeros, and test one point between consecutive zeros. Continuity guarantees the sign is constant across each region, so a single test settles it.
Whether an endpoint is included depends on the inequality: a strict inequality excludes every zero, while a non-strict one includes them. That decision is separate from the sign analysis and is made afterwards.
Figure (svg): A sign diagram for two sine x minus one on the interval from zero to two pi, showing the expression negative before pi over six, positive between pi over six and five pi over six, and negative afterwards
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 866-866
Picture it
Both break the line into regions. Only one of them can ever be part of the answer.
Figure (svg): A summary of the two kinds of landmark on a sign diagram, the zeros of the expression and the points where it is undefined, with the note that only zeros can be included in an answer
An undefined point is excluded no matter what the inequality sign says, because the expression does not exist there and cannot satisfy anything.
Worked example
Example 10.7.3, part 1. Three regions, three test values.
\[ \text{Solve } 2\sin(x) \le 1 \text{ on } [0, 2\pi). \]
Move everything to one side
Why: Compare against zero and name the left side f.
\[ f(x) = 2 \sin x - 1 \le 0 \]
Find the landmarks
Why: The sine is defined everywhere, so only zeros matter.
\[ x = \frac{\pi}{6}\text{ and } 5 \pi / 6 \]
Test one point per region
Why: Zero, pi over two and pi are convenient.
\[ f = -1, +1, -1 \]
Collect the regions with the sign wanted
Why: Negative or zero, and the inequality is non-strict so the zeros are included.
\[ [0, \frac{\pi}{6}]\text{ and } [5 \pi / 6, 2 \pi] \]
Figure (svg): The solution to Worked example a basic inequality shown as a ladder of expressions, one row per legal move
\[ \left[0, \tfrac{\pi}{6}\right] \cup \left[\tfrac{5\pi}{6}, 2\pi\right) \]
Verify: check an interior point of each interval
Why: At x equal to zero, twice the sine is 0, which is at most 1. At x equal to three pi over two, twice the sine is negative 2, also at most 1. And at pi over two, twice the sine is 2, which is not — correctly excluded.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 866-866
Sorting
A landmark is a zero of the expression or a point where it is undefined.
Sort into buckets
For f of x equal to tangent x minus 1 on the interval from zero to two pi, sort each point.
Worked example
Example 10.7.3, part 2. Move everything over and factor, exactly as for an equation.
\[ \text{Solve } \sin(2x) > \cos(x) \text{ on } [0, 2\pi). \]
Move everything to one side
Why: Do not divide by the cosine.
\[ \sin 2 x - \cos x > 0 \]
Factor after the double angle identity
Why: Two sine x cosine x minus cosine x.
\[ \cos x(2 \sin x - 1) > 0 \]
Find the zeros
Why: Each factor vanishing gives landmarks.
\[ \frac{\pi}{6}, \frac{\pi}{2}, 5 \pi / 6, 3 \pi / 2 \]
Test each of the five regions
Why: Zero, pi over four, three pi over four, pi and seven pi over four.
\[ -, +, -, +, - \]
Figure (svg): The solution to Worked example an inequality with two functions shown as a ladder of expressions, one row per legal move
\[ \left(\tfrac{\pi}{6}, \tfrac{\pi}{2}\right) \cup \left(\tfrac{5\pi}{6}, \tfrac{3\pi}{2}\right) \]
Verify: check a point in each answer interval
Why: At x equal to pi over four, the sine of pi over two is 1 and the cosine of pi over four is about 0.707, so the inequality holds. At x equal to pi, the sine of two pi is 0 and the cosine of pi is negative 1, so it holds there too. Both intervals check, and the inequality is strict so every endpoint is excluded.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 867-867
Trap
\[ \sin(2x) > \cos(x) \;\Longrightarrow\; 2\sin(x)\cos(x) > \cos(x) \;\Longrightarrow\; 2\sin(x) > 1 \]
Divide both sides by the cosine
Why: It appears on both sides, and cancelling looks like a simplification.
Two things go wrong at once. Solutions where the cosine is zero are lost, and where the cosine is negative the inequality sign should have reversed — which it did not.
\[ \cos(x)\big(2\sin(x) - 1\big) > 0 \]
Move everything to one side, factor, and use a sign diagram
Why: No division, so no lost solutions and no sign question.
Dividing an inequality by an expression of unknown sign is worse than dividing an equation, because it can silently flip the direction as well as discard solutions. The rule is therefore even more absolute here: compare against zero and make a diagram.
Fill the middle
Solve cosine x greater than one half on the interval from zero to two pi.
Fill in the blanks
\cos x - \tfracpi/35 pi/3 > 0, \text___ ___ \text___ ___ \;\Longrightarrow\; \left[0, \tfrac______\right) \cup \left(\tfrac______, 2\pi\right)
Why: The cosine equals one half at pi over three and five pi over three. Testing at zero gives a positive value, between them gives negative, and past five pi over three gives positive again. The answer is the two outer regions, with the zeros excluded because the inequality is strict — but zero and the approach to two pi are included, since those are the window's own endpoints rather than zeros of the expression.
Prediction
An expression has three zeros inside a stated interval and is defined everywhere.
Predict first
How many test values do you need?
Correct: Four.
Why: Three zeros cut the interval into four regions, and one test value per region settles the sign throughout it by continuity. In general n landmarks give n plus one regions. Counting the regions before testing is a useful check that none has been overlooked, particularly when some landmarks are undefined points rather than zeros.
Socratic
The method insists on comparing against zero rather than against the original right-hand side.
Discussion prompt
Explain why zero is special here, and what would go wrong comparing two nonzero expressions directly.
Hint: What do you know about the sign of a number that is not zero?
Answer:
Zero is the only number whose sign is unambiguous — everything else is either positive or negative and the comparison depends on which. Asking whether f of x is positive has a definite answer at each point; asking whether f of x exceeds g of x requires knowing both.
Comparing two expressions directly also loses the continuity argument. The claim that the sign is constant between consecutive zeros is a claim about one function, and it is what makes a single test value sufficient. With two expressions there is no single function to apply it to, until you subtract and make one.
Section
Section 2
Concept
An expression involving tangent, cotangent, secant or cosecant has undefined points as well as zeros, and both break the diagram. And when a zero is an inverse-function value rather than a familiar angle, choosing a test value beside it needs a little thought.
The same trick applies to any increasing function with an asymptote. It is not a special dodge so much as the general observation that a monotone function preserves order.
Figure (svg): A sign diagram for tangent x minus three on the interval from zero to two pi, with zeros at the arctangent of three and pi plus that, and undefined points at pi over two and three pi over two
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 867-868
Picture it
The gap between the arctangent of 3 and pi over two contains no familiar angle at all.
Figure (svg): The trick for choosing a test value between the arctangent of three and pi over two, using the arctangent of a much larger number which is guaranteed to lie in that gap
Any value between them will do, and the arctangent of 117 is both guaranteed to be in the gap and trivially easy to substitute, since the tangent undoes it immediately.
Worked example
Example 10.7.3, part 3. Five regions, two of them bounded by asymptotes.
\[ \text{Solve } \tan(x) \ge 3 \text{ on } [0, 2\pi). \]
Move everything over and name f
Why: Compare against zero.
\[ f(x) = \tan x - 3 \ge 0 \]
Find the undefined points
Why: The tangent fails at the odd multiples of pi over two.
\[ \frac{\pi}{2}\text{ and } 3 \pi / 2 \]
Find the zeros
Why: The tangent equals 3, which is not a special value.
\[ \arctan 3\text{ and } \arctan 3 + \pi \]
Test each region and collect
Why: Five regions; the two containing the zeros' right sides are positive.
Figure (svg): The solution to Worked example an inequality with undefined points shown as a ladder of expressions, one row per legal move
\[ \left[\arctan(3), \tfrac{\pi}{2}\right) \cup \left[\arctan(3) + \pi, \tfrac{3\pi}{2}\right) \]
Verify: check the endpoint types
Why: The zeros are included because the inequality is non-strict, so square brackets. The asymptotes are excluded because the tangent does not exist there, so round brackets — and that would be true even if the inequality were non-strict, which is the whole point of distinguishing the two landmark types.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 867-868
Discrimination
Zeros follow the inequality sign; undefined points never do.
Sort into buckets
For the inequality f of x at least zero, sort each landmark type.
Worked example
The gap between the arctangent of 3 and pi over two, from the same problem.
\[ \text{Find a usable test value in } \left(\arctan(3), \tfrac{\pi}{2}\right) \text{ and evaluate } \tan(x) - 3 \text{ there.} \]
Note the arctangent is increasing
Why: A larger input gives a larger output.
Note it is bounded above by pi over two
Why: That bound is its horizontal asymptote.
\[ \text{always below } \frac{\pi}{2} \]
Pick any number larger than 3
Why: One hundred seventeen will do; so would 4.
\[ \arctan(117) \]
Evaluate the expression there
Why: The tangent undoes the arctangent immediately.
\[ 117 - 3 = 114 > 0 \]
Figure (svg): The solution to Worked example choosing the awkward test value shown as a ladder of expressions, one row per legal move
\[ f\big(\arctan(117)\big) = 117 - 3 = 114 > 0 \]
Verify: confirm the value is in the gap
Why: Since 117 is greater than 3 and the arctangent is increasing, its arctangent exceeds the arctangent of 3. And every arctangent is below pi over two. So the value is strictly between the two landmarks, which is all that was required.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 867-868
Error analysis
A student solves tangent x at least 3 and writes the answer.
Annotate
On: \( \left[\arctan(3), \tfrac{\pi}{2}\right] \cup \left[\arctan(3) + \pi, \tfrac{3\pi}{2}\right] \)
The inequality sign decides only what happens at the zeros. Undefined points are excluded by the fact that the expression does not exist, which no inequality sign can override.
Prediction
You need a test value between the arctangent of 5 and pi over two.
Predict first
Which of these is guaranteed to lie there?
Correct: arctan(50).
Why: Since the arctangent is increasing and 50 exceeds 5, its arctangent exceeds the arctangent of 5; and every arctangent is below pi over two. So arctan(50) is guaranteed to be in the gap. The arctangent of 2 is smaller, not larger. Pi over three is about 1.047 while the arctangent of 5 is about 1.373, so it falls short. And 1.6 exceeds pi over two, about 1.571, so it is past the gap entirely.
Faded example
For f of x equal to cotangent x minus 1 on the interval from zero to pi, identify the landmarks.
Fill in the blanks
\text0 x = pi/4 \text___ x = \pi; \quad \text___ x = ___
Why: The cotangent is undefined at the multiples of pi, which includes both endpoints of this interval, and it equals 1 at pi over four. So the open interval is cut into two regions by a single zero. Testing at pi over six gives a positive value and at pi over two gives negative one, so the expression is positive on the first region and negative on the second.
Edge cases
The sign diagram method relies on the expression being continuous between landmarks.
Discussion prompt
Is a trigonometric expression continuous across an asymptote? What would go wrong if you tested only one value on each side of a zero and ignored the asymptotes?
Hint: Can a function change sign without passing through zero?
Answer:
It is not continuous across an asymptote — that is what an asymptote is. So a function can and does change sign there without passing through zero, which is exactly what the tangent does at pi over two.
Ignoring the asymptotes would mean treating the region from the arctangent of 3 all the way to the arctangent of 3 plus pi as a single region, and one test value in it would give the wrong sign for most of it. Undefined points must be marked as landmarks precisely because the continuity argument fails there, and that is the whole reason they appear on the diagram at all.
Section
Section 3
Concept
A function built from circular functions is undefined wherever a denominator vanishes. Finding those points is an ordinary trigonometric equation, and the domain is everything else.
The book's advice is worth repeating: when in doubt, write it out. Listing six or eight excluded points on a number line makes the interval structure obvious in a way the formula alone does not.
Figure (svg): The two sources of a domain restriction for a function built from circular functions: a denominator that vanishes and a square root of something negative
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 868-869
Picture it
A denominator gives an equation to solve; a radical gives an inequality.
Figure (svg): The two sources of a domain restriction for a function built from circular functions: a denominator that vanishes and a square root of something negative
That is why this material sits in the same lesson as the inequalities: half of every domain question is an inequality problem, and the other half is an equation problem.
Worked example
Example 10.7.4, part 1. Rewrite as a fraction and find the denominator's zeros.
\[ \text{Find the domain of } f(x) = \csc\left(2x + \frac{\pi}{3}\right). \]
Rewrite in terms of sine
Why: The cosecant is one over the sine.
\[ f = 1 / \sin(2 x + \frac{\pi}{3}) \]
Set the denominator to zero
Why: The sine vanishes at multiples of pi.
\[ 2 x + \frac{\pi}{3} = \pi k \]
Solve for x
Why: Subtract and halve, dividing the step too.
\[ x = -\frac{\pi}{6} + (\frac{\pi}{2}) k \]
Describe the domain
Why: Everything except those points, spaced pi over two apart.
Figure (svg): The solution to Worked example the domain of a cosecant shown as a ladder of expressions, one row per legal move
\[ \text{dom}(f) = \left\{x : x \ne -\tfrac{\pi}{6} + \tfrac{\pi}{2}k, \; k \in \mathbb{Z}\right\} \]
Verify: write out several
Why: Taking k from negative one to two gives negative two pi over three, negative pi over six, pi over three and five pi over six. Consecutive excluded points differ by pi over two throughout, which matches the step — and that spacing is half the period of sine of 2x plus pi over three, as it should be, since a sine vanishes twice per period.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 868-868
Matching
Rewrite in terms of sine and cosine, then find the denominator's zeros.
Match the pairs
Why: The first two are the standard exclusions from Lesson 10.3a. The last two need the denominator solved: the sine equals 1 only at pi over two plus two pi k, a single family since the line is tangent there, and the cosine equals negative one only at pi plus two pi k for the same reason. Both are single families rather than two, which is the exceptional case for a sine or cosine equation.
Worked example
Example 10.7.4, part 2. Only the denominator matters.
\[ \text{Find the domain of } f(x) = \frac{\sin(x)}{2\cos(x) - 1}. \]
Note the numerator causes no trouble
Why: The sine is defined for every real number.
Set the denominator to zero
Why: Two cosine x minus one equals zero.
\[ \cos x = \frac{1}{2} \]
Solve
Why: Two families, since a cosine equation has two.
\[ x = \frac{\pi}{3} + 2 \pi k\text{ or } 5 \pi / 3 + 2 \pi k \]
Describe the domain
Why: Everything except those two families.
Figure (svg): The solution to Worked example a rational function of circular functions shown as a ladder of expressions, one row per legal move
\[ \text{dom}(f) = \left\{x : x \ne \tfrac{\pi}{3} + 2\pi k \;\text{ and }\; x \ne \tfrac{5\pi}{3} + 2\pi k\right\} \]
Verify: write out the pattern
Why: The excluded points are plus and minus pi over three, plus and minus five pi over three, plus and minus seven pi over three, and so on. They are not evenly spaced: the gaps alternate between two pi over three and four pi over three, which is why two families are needed to describe them rather than one.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 869-869
Trap
\[ \text{excluded: } x = \tfrac{\pi}{3} + \tfrac{2\pi}{3}k \]
Notice the first gap is two pi over three and assume the spacing is uniform
Why: The first two excluded points really are that far apart, so the pattern looks established.
But the next gap is four pi over three, not two pi over three. The formula would exclude pi, where the cosine is negative one and the denominator is perfectly fine.
\[ x \ne \tfrac{\pi}{3} + 2\pi k \quad\text{and}\quad x \ne \tfrac{5\pi}{3} + 2\pi k \]
Keep the two families separate, as the equation produced them
Why: A cosine equation gives two families whose members interleave unevenly.
This is the same fact as in Lesson 10.2b: cosine and sine families do not combine, because their two solution quadrants are adjacent rather than opposite. Writing out six or eight excluded points and measuring the gaps is the cheapest way to see whether a single formula can work.
Faded example
Find the domain of one over the quantity tangent x minus 1.
Fill in the blanks
\textpi/2 + pi k \tan x \textpi/4 + pi k x = ___; \quad \text___ \tan x = 1: x = ___
Why: Two separate sources of exclusion. The tangent itself fails at the odd multiples of pi over two, and the denominator vanishes wherever the tangent equals 1, at pi over four plus pi k. Both sets must be removed, and forgetting the first is the commonest omission — the function inside the expression has its own domain before the fraction is even considered.
Prediction
You are finding the domain of a function whose denominator is 2 sine x minus 3.
Predict first
What will you have to exclude?
Correct: Nothing at all.
Why: Setting the denominator to zero gives a sine of three halves, which is impossible since the sine never exceeds 1. So the denominator never vanishes and nothing is excluded — the domain is all real numbers. Checking the constant against the range before solving is worth doing here as much as anywhere else, and it turns the whole question into one line.
Notation
This is extended interval notation for a domain. Read what each part is doing.
Annotate
On: \( \bigcup_{k=-\infty}^{\infty} \left( \frac{(3k-1)\pi}{6}, \; \frac{(3k+2)\pi}{6} \right) \)
The notation is heavy and the set-builder form usually reads better. It is worth being able to decode because it appears in the book and in calculus texts, but writing the domain as everything except a stated family is generally clearer.
Section
Section 4
Concept
A square root demands that its contents be non-negative, which is an inequality. When no interval is stated, solve it over one period of the function involved and then translate by every multiple of that period.
This works only because the function is periodic: the sign pattern over one period is the sign pattern over every period, so one diagram genuinely does describe the whole line.
Figure (svg): The strategy of solving an inequality on one fundamental period and then adding integer multiples of that period to describe the whole solution set
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 870-870
Picture it
The shaded band is the solution on one period; every other band is a translate of it.
Figure (svg): The strategy of solving an inequality on one fundamental period and then adding integer multiples of that period to describe the whole solution set
The translation step is what makes an unbounded problem tractable, and periodicity is what makes the translation legitimate.
Worked example
Example 10.7.4, part 3. Two restrictions, one of them an inequality.
\[ \text{Find the domain of } f(x) = \sqrt{1 - \cot(x)}. \]
Impose the cotangent's own domain
Why: It is undefined at multiples of pi.
Impose the radical's condition
Why: The contents must be at least zero.
\[ 1 - \cot x \ge 0 \]
Solve on one period
Why: Cotangent has period pi, so use the interval from zero to pi.
\[ \text{zero at } \frac{\pi}{4} \]
Test and translate
Why: Testing pi over six and pi over two gives negative then positive.
\[ [\frac{\pi}{4}, \pi]\text{ then add } \pi k \]
Figure (svg): The solution to Worked example the domain of a radical shown as a ladder of expressions, one row per legal move
\[ \bigcup_{k=-\infty}^{\infty} \left[\tfrac{(4k+1)\pi}{4}, \; (k+1)\pi\right) \]
Verify: check a value in one interval
Why: At x equal to pi over two the cotangent is zero, so the radicand is 1, which is non-negative — the point is in the domain, and it lies in the interval from pi over four to pi. At x equal to pi over six the cotangent is root three, so the radicand is 1 minus root three, negative — correctly excluded.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 870-870
Sorting
Different obstacles need different tools.
Sort into buckets
Sort each feature by what it requires.
Worked example
The same method with a period of two pi rather than pi.
\[ \text{Find the domain of } g(x) = \sqrt{2\sin(x) - 1}. \]
Impose the radical's condition
Why: The contents must be non-negative.
\[ 2 \sin x - 1 \ge 0 \]
Solve on one period
Why: The sine has period two pi, so use zero to two pi.
\[ \text{zeros at } \frac{\pi}{6}\text{ and } 5 \pi / 6 \]
Test the three regions
Why: Zero, pi over two and pi give negative, positive, negative.
Translate by the period
Why: Add two pi k to both endpoints.
\[ [\frac{\pi}{6} + 2 \pi k, 5 \pi / 6 + 2 \pi k] \]
Figure (svg): The solution to Worked example a radical with a sine inside shown as a ladder of expressions, one row per legal move
\[ \bigcup_{k=-\infty}^{\infty} \left[\tfrac{\pi}{6} + 2\pi k, \; \tfrac{5\pi}{6} + 2\pi k\right] \]
Verify: check the endpoints are included
Why: At pi over six the sine is one half, so the radicand is exactly zero and the square root is defined, giving zero. Non-negative includes zero, so both endpoints belong and the brackets are square. Contrast the previous example, where the right endpoint was excluded because the cotangent was undefined there rather than because the radicand was negative.
Error analysis
A student finds the domain of the square root of one minus cotangent x.
Annotate
On: \( 1 - \cot(x) \ge 0 \;\Longrightarrow\; \left[\tfrac{\pi}{4} + \pi k, \; \pi + \pi k\right] \)
A composite function inherits every restriction from every part of it. Listing the sources separately — the inner function's domain, then the radical, then any denominator — before combining them prevents this.
Faded example
Find the domain of the square root of cosine x.
Fill in the blanks
\cos x \ge 0 \textpi/2 [0, 2\pi) \text3 pi/2 \left[0, ___\right] \cup \left[___, 2\pi\right), \text___ 2\pi k
Why: The cosine is non-negative on the right half of the circle, which within one revolution is from zero to pi over two and again from three pi over two to two pi. Those two pieces are in fact adjacent once the periodicity is taken into account, so the domain is more compactly written as the union of intervals from negative pi over two plus two pi k to pi over two plus two pi k.
Prediction
You solve an inequality involving the tangent over one period and get one interval.
Predict first
How many intervals will the full solution set contain?
Correct: Infinitely many.
Why: Periodicity means the solution repeats in every period, and there are infinitely many periods on the real line. So one interval per period gives infinitely many intervals, all translates of the first. This is why extended interval notation exists: describing infinitely many intervals compactly requires a union indexed by the integers, and writing them out is not an option.
Explain it to yourself
Solving on one period and translating is presented as a strategy rather than a theorem.
Discussion prompt
Explain why it is legitimate, and identify the property of the function that makes it work.
Hint: What does periodicity say about the values at x and at x plus the period?
Answer:
Periodicity says the function takes exactly the same value at x and at x plus the period. So if the inequality holds at some point, it holds at that point shifted by any whole number of periods, and if it fails somewhere, it fails at all the translates too.
That means the solution set is invariant under translation by the period, so knowing it on any one period determines it everywhere. The property doing the work is periodicity itself, and the strategy would fail immediately for a non-periodic function — for x squared minus one, say, where solving on any single interval says nothing about the rest.
Section
Section 5
Concept
An equation such as arcsine of 2x equals pi over three looks like the equations of the previous lesson and behaves in the opposite way. Because an inverse function is one-to-one, such an equation has at most one solution.
The cancellation used is the outside-in direction, which always works on the inverse's domain. That is why this is straightforward, whereas the inside-out direction would need a condition checked.
Figure (svg): Two columns contrasting an equation in a circular function with one in an inverse circular function
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 870-871
Picture it
They look alike and behave oppositely, which is worth having clearly in view.
Figure (svg): Two columns contrasting an equation in a circular function with one in an inverse circular function
An equation in a circular function has infinitely many solutions because the function is periodic. An equation in an inverse has at most one because the inverse is one-to-one.
Worked example
Example 10.7.5, part 1. Apply the sine to both sides.
\[ \text{Solve } \arcsin(2x) = \frac{\pi}{3}. \]
Check the right side is in range
Why: Pi over three is between negative pi over two and pi over two.
Apply the sine to both sides
Why: The outside-in cancellation always works.
\[ 2 x = \sin(\frac{\pi}{3}) \]
Evaluate
Why: The sine of pi over three is root three over two.
\[ 2 x = \sqrt{3} / 2 \]
Solve
Why: Divide by 2.
\[ x = \sqrt{3} / 4 \]
Figure (svg): The solution to Worked example an arcsine equation shown as a ladder of expressions, one row per legal move
\[ x = \frac{\sqrt{3}}{4} \approx 0.4330 \]
Verify: substitute back
Why: Twice root three over four is root three over two, and the arcsine of that is pi over three — since pi over three is in the arcsine's range. And the answer is the only one, because the arcsine is one-to-one.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 870-871
Sorting
Check the right side against the range of the inverse function.
Sort into buckets
Sort each equation.
Worked example
Example 10.7.5, part 2. One algebraic step before the cancellation.
\[ \text{Solve } 4\arccos(x) - 3\pi = 0. \]
Isolate the arccosine
Why: Add three pi and divide by 4.
\[ \arccos x = 3 \pi / 4 \]
Check the range
Why: Three pi over four is between zero and pi.
Apply the cosine to both sides
Why: The cancellation is valid on the arccosine's domain.
\[ x = \cos(3 \pi / 4) \]
Evaluate
Why: The cosine of three pi over four is negative root two over two.
\[ x = -\sqrt{2} / 2 \]
Figure (svg): The solution to Worked example isolate first shown as a ladder of expressions, one row per legal move
\[ x = -\frac{\sqrt{2}}{2} \approx -0.7071 \]
Verify: substitute back
Why: The arccosine of negative root two over two is three pi over four, and four times that minus three pi is three pi minus three pi, which is zero. The answer checks, and again there is exactly one because the arccosine is one-to-one.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 871-871
Trap
\[ \arcsin(x) = 2 \;\Longrightarrow\; x = \sin(2) \approx 0.909 \]
Apply the sine to both sides and evaluate
Why: The manipulation is the standard one and it produces a number.
But 2 is greater than pi over two, about 1.571, so it is outside the range of the arcsine. No input produces an arcsine of 2, so the equation has no solutions at all.
Check first: is the right-hand side in the range of the inverse function? Here it is not, so the answer is no solutions, and no further work is needed.
\[ \text{range}(\arcsin) = \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right], \quad 2 \notin \text{that interval} \]
Applying the sine anyway produces a number that fails when substituted back: the arcsine of 0.909 is about 1.14, not 2. The manipulation is not reversible when the range check fails, which is exactly why the check comes first.
Faded example
Solve 3 arcsecant of the quantity 2x minus 1, plus pi, equals 2 pi.
Fill in the blanks
\operatornamepi/3(2x - 1) = 2 \;\Longrightarrow\; 2x - 1 = \sec\left(___\right) = ___ \;\Longrightarrow\; x = \tfrac______
Why: Isolating gives an arcsecant of pi over three. That value lies in both of the competing arcsecant ranges, so the convention does not matter here and the cancellation is safe either way. Applying the secant gives 2x minus 1 equal to 2, so x is three halves. The book notes explicitly that both ranges contain pi over three, which is why this problem is convention-free.
Prediction
You are solving an equation in which the only unknown appears inside an arccosine.
Predict first
How many solutions can it have?
Correct: At most one.
Why: The arccosine is one-to-one, so it takes each value at most once, and an equation setting it equal to a constant therefore has at most one solution. It has none when the constant is outside the range. This is the exact opposite of an equation in the cosine itself, which is periodic and so has infinitely many solutions whenever it has any — and confusing the two situations is the commonest error in this material.
Counterexample
A student proposes: since the cosine and the arccosine are inverses, an equation in one has the same number of solutions as the corresponding equation in the other.
Discussion prompt
Give a specific pair of equations that makes this false, and explain the structural reason.
Hint: Compare cosine x equals one half with arccosine x equals pi over three.
Answer:
\[ \cos(x) = \tfrac{1}{2}: \text{ infinitely many solutions} \]
\[ \arccos(x) = \tfrac{\pi}{3}: \text{ exactly one, } x = \tfrac{1}{2} \]
The structural reason is that the cosine is periodic and the arccosine is one-to-one. Periodicity means a value is achieved infinitely often; being one-to-one means a value is achieved at most once.
The restriction that created the arccosine is precisely what destroyed the periodicity, so the two functions could not possibly behave the same way. The inverse of a many-to-one function is one-to-one by construction, and that asymmetry is the whole content of Section 10.6.
Comparison
Fill the blanks from memory. The setup is shared; the finish is not.
Comparison matrix
| Equation | Inequality | |
|---|---|---|
| First move | everything to one side | everything to one side |
| Then | solve f(x) = 0 | find the landmarks, then test |
| Answer is | a set of points | a union of intervals |
| Undefined points | irrelevant | landmarks, always excluded |
| Never do | divide by an expression | divide by an expression |
Solving the equation is the first half of solving the inequality, which is why this lesson could not have come first.
Pattern
Whether the question is an inequality or a domain, the same five moves cover it.
For an equation or inequality in an inverse trigonometric function, none of this applies. Check the constant against the range, apply the corresponding function to both sides, and expect at most one solution.
OpenStax Algebra and Trigonometry 2e, §9.5 Solving Trigonometric Equations §9.5
Check
A basic inequality. Watch the endpoints.
Check your understanding
Solve cosine x at most zero on the interval from zero to two pi.
Answer: B
Why: The cosine is zero at pi over two and three pi over two, and negative between them. Since the inequality is non-strict, the zeros are included, giving the closed interval. Testing at pi confirms the middle region: the cosine there is negative one, which is at most zero.
Check
Endpoints again, this time at an asymptote.
Check your understanding
The solution of an inequality includes the region up to a point where the expression is undefined. What bracket does that endpoint get?
Answer: B
Why: An undefined point cannot satisfy any inequality, because the expression has no value there to compare. So it is excluded regardless of whether the inequality is strict, and the bracket is always round.
Check
An inverse equation. Check the range first.
Check your understanding
How many solutions does arccos x equal 5 pi over 4 have?
Answer: A
Why: The range of the arccosine is the interval from zero to pi, and five pi over four is about 3.93 while pi is about 3.14. So the required value is outside the range and no input produces it. The equation has no solutions.
Real world
A wind turbine only generates usefully when the wind speed exceeds 4 metres per second. At a coastal site the daily wind speed is modelled as 6 plus 3 times the sine of the quantity pi over 12 times t, with t in hours after midnight.
Discussion prompt
Determine the hours of the day during which the turbine generates, and say why the answer is an interval rather than a set of times.
Hint: Set up the inequality and use a sign diagram over one period.
Answer:
\[ 6 + 3\sin\left(\tfrac{\pi}{12}t\right) > 4 \;\Longrightarrow\; \sin\left(\tfrac{\pi}{12}t\right) > -\tfrac{2}{3} \]
The zeros come from the sine equal to negative two thirds, giving the argument as arcsine of negative two thirds plus two pi k, or pi minus that. Unwinding gives t about 18.79 and t about 29.21, and reducing the second into the day gives about 5.21.
So the turbine generates from about 05:13 until about 18:47 — roughly 13.6 hours — and is idle overnight. The answer is an interval because the condition is about exceeding a threshold, which happens over a continuous stretch of time rather than at isolated instants.
That distinction is the whole reason inequalities are worth a separate lesson. An equation would have told you the two moments when the wind speed is exactly 4; the inequality tells you the period of operation, which is what the operator actually needs.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
How many solutions can the equation arctan x equals c have?
Correct: At most one, and none if c is outside the range.
\[ \arctan(x) = \tfrac{\pi}{2}: \text{ no solutions, since } \tfrac{\pi}{2} \notin \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right) \]
Why: The arctangent is one-to-one, so it takes each value at most once and the equation cannot have two solutions. It has exactly one when c lies strictly between negative pi over two and pi over two, and none otherwise — for instance arctan x equals pi over two has no solution, because pi over two is an asymptote the function approaches without reaching. The second option is nearly right and fails precisely at those excluded values, which is why the range check is not optional.
Explain it
They can solve trigonometric equations and are asked to solve an inequality for the first time. They want to move terms across the inequality sign the way they would in an equation.
Discussion prompt
In no more than five sentences, tell them what to do instead and why the shortcut fails.
Hint: What operation on an inequality is dangerous, and why is a trigonometric expression especially bad?
Answer:
A usable answer: adding and subtracting is fine, but dividing by something whose sign you do not know is not — because a negative divisor flips the inequality and you cannot know when that happens. A trigonometric expression is positive on some intervals and negative on others, so it is exactly the worst case.
Instead, move everything to one side so you are comparing against zero, find every point where the expression is zero or undefined, and test one value between consecutive such points. The sign cannot change in between, so one test settles each region — and the answer is the regions you want, written as intervals.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Sign diagrams are fixed by listing both kinds of landmark before testing anything. Endpoints are fixed by one rule: zeros follow the inequality sign, undefined points never do. Radical domains are fixed by solving the inequality on one period and translating. Inverse equations are fixed by checking the constant against the range before applying anything, and remembering there is at most one solution. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page draw a number line from zero to two pi and build the complete sign diagram for tangent x minus 1, marking zeros with one symbol and undefined points with another, writing a test value and its sign in each region, and stating the solution of the inequality tangent x at least 1. Below, write the two-line rule for endpoints: what happens at a zero and what happens at an undefined point. In the middle right, write the two sources of a domain restriction and which tool each needs. Underneath, find the domain of the square root of one minus two cosine x completely, solving on one period and then writing the translated answer. In the bottom left, solve two arcsin x minus pi equals zero, showing the range check. In the bottom right, write one sentence contrasting how many solutions an equation in sine has with how many an equation in arcsine has, and say what causes the difference. Finally, circle the one operation on the page that is forbidden throughout.
The circled operation is dividing both sides by a trigonometric expression. For an equation it loses solutions; for an inequality it can also flip the sign, which makes it worse rather than merely incomplete.
Recap
Five things, and they close out Chapter 10.
| If the question says | Your first move is |
|---|---|
| Solve this inequality | Move everything to one side and name it f |
| Find the domain | List every denominator and every even radical |
| No interval is given | Solve on one period, then translate |
| An inverse function appears | Check the constant against its range |
| A test value is awkward | Use monotonicity to find one in the gap |
That completes Chapter 10. Every one of the six functions has been defined, computed, graphed, inverted and solved. Chapter 11 turns outward: the same functions applied to triangles that are not right-angled, to a second coordinate system, to complex numbers, and to vectors — six genuinely different subjects, all resting on what this chapter built.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.7 Trigonometric Equations and Inequalities §10.7, pp. 866-874 — everything on these slides traces back here
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