How do you solve a trigonometric inequality?

First solve the corresponding equation to find the boundary angles where equality holds. Then determine which intervals between these boundary points satisfy the inequality, using either the unit circle or the graph of the function. Express the solution as a union of intervals, and add the period to capture all solutions if a general solution is needed.

What is the graphical method for solving trig inequalities?

Graph the trigonometric function and the constant value on the same axes. The inequality is satisfied wherever the function's graph lies above (for >) or below (for <) the horizontal line. Read the solution intervals directly from the graph by identifying the intersection points and the regions that satisfy the condition.

How do you use the unit circle to solve trig inequalities?

On the unit circle, sinθ corresponds to the y-coordinate and cosθ to the x-coordinate. For an inequality like sinθ > 1/2, draw the horizontal line y = 1/2 and identify the arc where the circle lies above it. The angles corresponding to that arc form the solution. This method gives a geometric view of which angles satisfy the inequality.

How do you express trigonometric inequality solutions?

Solutions within one period are written as open or closed intervals depending on whether the inequality is strict ( ) or non-strict (≤ or ≥). The general solution adds integer multiples of the period to each interval endpoint. Solutions are typically expressed in interval notation or set-builder notation.

What are compound trigonometric inequalities?

Compound trigonometric inequalities involve two bounds, such as -1/2 ≤ sinx ≤ √3/2. These are solved by finding the solution set for each individual inequality and then taking the intersection. The result is the set of angles where both conditions hold simultaneously, which can be read from the graph or unit circle.

Trigonometric Inequalities

Q: How do you use the unit circle to solve trig inequalities?

On the unit circle, sinθ corresponds to the y-coordinate and cosθ to the x-coordinate. For an inequality like sinθ > 1/2, draw the horizontal line y = 1/2 and identify the arc where the circle lies above it. The angles corresponding to that arc form the solution. This method gives a geometric view of which angles satisfy the inequality.

Key Terms

Solutions as Intervals

The Graphical Method

The Unit Circle Method

Solving Sine Inequalities

Solving Cosine and Tangent Inequalities

Compound Inequalities

Inequalities Involving Transformed Functions

Domain Considerations for Reciprocal and Quotient Functions

Summary of Inequality Forms and Their Solution Approaches

Finding Intervals Where Trigonometric Conditions Hold

A trigonometric equation asks: at which angles does a function hit a specific value? A trigonometric inequality asks a broader question: over which intervals does a function stay above, below, or between given bounds? The answer is not a set of isolated angles but a union of intervals — continuous stretches of the real line where the condition is satisfied. And because the trigonometric functions are periodic, these intervals repeat.

Two methods dominate. The graphical approach plots the function and a horizontal boundary line, then reads off the regions where the curve lies on the correct side. The unit circle approach identifies the boundary angles (the solutions of the corresponding equation) and determines which arc of the circle satisfies the inequality by examining the sign of the function along that arc. Both methods yield the same result; the choice depends on whether a visual or geometric perspective is more natural for the problem at hand. In either case, the underlying logic draws on the same properties — periodicity, boundedness, continuity — and the same evaluation skills used in equation solving.

Key Terms

Unit Circle— geometric tool for reading solution intervals from coordinates

Reference Angle— acute angle to the

x

-axis, used to locate boundary points

Periodic Function— repeating behavior that extends solution sets across periods

Quadrantal Angles— boundary angles where some functions are undefined

Amplitude— controls the vertical range of transformed inequalities

Period— determines how many solution intervals appear in a given domain

See All Trigonometry Definitions →

Solutions as Intervals

The solution to a trigonometric inequality is fundamentally different from the solution to an equation. An equation like

\sin(x) = \frac{1}{2}

produces isolated points:

x = \frac{\pi}{6}, \frac{5\pi}{6}

, plus their periodic repetitions. An inequality like

\sin(x) > \frac{1}{2}

produces intervals: the entire stretch of angles between

\frac{\pi}{6}

and

\frac{5\pi}{6}

(exclusive), plus periodic repetitions.

This shift from points to intervals is a consequence of continuity. Since sine is continuous, if

\sin(a) > \frac{1}{2}

and

\sin(b) > \frac{1}{2}

and there is no point between

a

and

b

where

\sin(x) = \frac{1}{2}

, then

\sin(x) > \frac{1}{2}

for every

x

between

a

and

b

. The boundary points — the solutions of the corresponding equation — are where the function crosses the threshold, and the intervals between them are where the inequality holds.

On a restricted interval like

[0, 2\pi)

, the solution is a finite union of intervals. In general solution form, the pattern repeats: each interval shifts by

2\pi

(for sine, cosine, cosecant, secant) or

\pi

(for tangent, cotangent) to produce the complete solution set.

Strict inequalities (

>

<

) produce open intervals at the boundary points — the boundary is not included. Non-strict inequalities (

\geq

\leq

) produce closed intervals at boundary points — the boundary is included (provided the function is defined there). At vertical asymptotes, the interval is always open, since the function is undefined.

The two regimes — equation and inequality — contrast on every structural attribute that matters for how the solution is built and written, as the comparison below makes explicit.

Aspect	Trigonometric equation	Trigonometric inequality
Solution form	isolated points	intervals (or unions of intervals)
Boundary points	the solution itself	endpoints of solution intervals
Endpoint inclusion	not a distinction (equality <em>is</em> the answer)	open if strict (>, <); closed if non-strict (≥, ≤)
Asymptote handling	function defined at every solution	always excluded — function undefined there
General-solution form	x = a + n·period	(a, b) + n·period (and unions)

The Graphical Method

The graphical method is the most intuitive approach to trigonometric inequalities. It reduces the problem to a visual reading.

Step 1: Graph the function. Sketch or visualize

y = \sin(x)

(or whichever function appears in the inequality) over the relevant interval. The key features — zeros, extrema, asymptotes, period — should be familiar from the study of graphs.

Step 2: Draw the boundary line. Plot the horizontal line

y = a

, where

a

is the value on the right side of the inequality.

Step 3: Identify the intersection points. These are the solutions of the corresponding equation (

\sin(x) = a

), which can be found using the techniques from equation solving.

Step 4: Read the solution intervals. For

\sin(x) > a

: the intervals where the curve lies above the horizontal line. For

\sin(x) < a

: the intervals where the curve lies below.

Example: Solve

\cos(x) > \frac{1}{2}

[0, 2\pi)

.

The equation

\cos(x) = \frac{1}{2}

has solutions

x = \frac{\pi}{3}

and

x = \frac{5\pi}{3}

on this interval. The cosine graph starts at

1

(above

\frac{1}{2}

), drops to

\frac{1}{2}

\frac{\pi}{3}

, continues below

\frac{1}{2}

through the middle of the interval, rises back to

\frac{1}{2}

\frac{5\pi}{3}

, and returns to

1

2\pi

. The curve is above the line

y = \frac{1}{2}

on:

x \in \left[0, \frac{\pi}{3}\right) \cup \left(\frac{5\pi}{3}, 2\pi\right)

The endpoints

\frac{\pi}{3}

and

\frac{5\pi}{3}

are excluded because the inequality is strict (

>

, not

\geq

). The endpoint

0

is included because the interval

[0, 2\pi)

starts there and

\cos(0) = 1 > \frac{1}{2}

.

The graphical method is especially useful for transformed functions —

2\sin(3x) - 1 > 0

is harder to handle algebraically, but sketching the transformed graph and reading off the solution intervals is straightforward once the transformations are understood.

The Unit Circle Method

The unit circle provides a geometric alternative to the graphical method, particularly effective for basic inequalities involving

\sin(x)

and

\cos(x)

.

For sine inequalities: since

\sin\theta

is the

y

-coordinate on the unit circle, the inequality

\sin(x) > a

asks: for which angles is the

y

-coordinate above the line

y = a

?

Draw the horizontal line

y = a

across the unit circle. It intersects the circle at two points (assuming

|a| < 1

), dividing the circle into two arcs. The arc where points have

y

-coordinates above

a

is the solution. Reading off the angles at the intersection points gives the interval boundaries.

For cosine inequalities: since

\cos\theta

is the

x

-coordinate, the inequality

\cos(x) > a

asks: for which angles is the

x

-coordinate to the right of the vertical line

x = a

?

Draw the vertical line

x = a

across the unit circle. The arc to the right of this line gives the solution.

Example: Solve

\sin(x) \geq \frac{\sqrt{2}}{2}

[0, 2\pi)

.

On the unit circle, the horizontal line

y = \frac{\sqrt{2}}{2}

intersects the circle at angles

\frac{\pi}{4}

and

\frac{3\pi}{4}

. The arc from

\frac{\pi}{4}

\frac{3\pi}{4}

(going counterclockwise through the top of the circle) lies above this line. Since the inequality includes equality (

\geq

), the endpoints are included:

x \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right]

For tangent inequalities: the unit circle method is less direct, since tangent does not correspond to a single coordinate. One approach is to interpret

\tan\theta

as the slope of the terminal side and identify which slopes satisfy the inequality. Alternatively, convert to

\frac{\sin\theta}{\cos\theta}

and reason about the signs and magnitudes separately. In practice, the graphical method is often simpler for tangent inequalities.

Set side by side, the two methods differ less in what they yield than in what they require and where each is most direct.

Method	What you draw	How you read the solution	Best applied to
Graphical	the function's curve plus the horizontal line y = a	intervals where the curve lies above (> a) or below (< a) the line	any inequality — especially transformed functions A·f(Bx+C)+D and tangent
Unit circle	a horizontal line y = a (for sine) or vertical line x = a (for cosine) cutting the circle	the arc on the satisfying side; boundary angles read off at intersection points	basic sin / cos inequalities; geometric intuition for boundary angles

Solving Sine Inequalities

The full procedure for solving

\sin(x) > a

(or

\geq

<

\leq

) combines the boundary equation with interval determination.

Case $|a| < 1$: The equation

\sin(x) = a

has two solutions per period:

x_1

and

x_2 = \pi - x_1

(where

x_1 = \arcsin(a)

). The sign of

a

determines the quadrants. Within one period

[0, 2\pi)

$\sin(x) > a$ when $a > 0$ : the solution is $(x_1, x_2)$ — the arc where sine exceeds $a$ , passing through the maximum.
$\sin(x) > a$ when $a < 0$ : the solution is $[0, x_2) \cup (x_1, 2\pi)$ where $x_1 = \pi - \arcsin(a)$ and $x_2 = \arcsin(a) + 2\pi$ — the larger arc that includes the maximum. More carefully: the two boundary angles are $\pi + |\arcsin(a)|$ and $2\pi - |\arcsin(a)|$ (in Quadrants III and IV), and sine exceeds $a$ on the arc that does not pass through the minimum.

Case $a = 1$:

\sin(x) \geq 1

is satisfied only at

x = \frac{\pi}{2} + 2n\pi

(a single point per period).

\sin(x) > 1

has no solution.

Case $a = -1$:

\sin(x) \leq -1

is satisfied only at

x = \frac{3\pi}{2} + 2n\pi

\sin(x) < -1

has no solution.

Case $|a| > 1$:

\sin(x) > a

when

a > 1

: no solution (sine never exceeds 1).

\sin(x) > a

when

a < -1

: all real numbers (sine is always greater than any value below

-1

These cases follow directly from the boundedness of sine. Checking whether

a

falls within

[-1, 1]

before proceeding prevents unnecessary computation.

All positions of

a

relative to the function's range — interior, edge, and outside — collect into the table below across all four inequality directions.

Range of a	sin(x) > a	sin(x) ≥ a	sin(x) < a	sin(x) ≤ a
−1 < a < 1	standard intervals, open endpoints	standard intervals, closed endpoints	standard intervals, open endpoints	standard intervals, closed endpoints
a = 1	no solution	x = π/2 + 2nπ only (isolated points)	all real x except π/2 + 2nπ	all real x
a = −1	all real x except 3π/2 + 2nπ	all real x	no solution	x = 3π/2 + 2nπ only (isolated points)
a > 1	no solution	no solution	all real x	all real x
a < −1	all real x	all real x	no solution	no solution

Solving Cosine and Tangent Inequalities

Cosine inequalities follow the same logic as sine, with the roles of the coordinates swapped. The equation

\cos(x) = a

(for

|a| < 1

) has solutions

x_1 = \arccos(a)

and

x_2 = 2\pi - \arccos(a)

[0, 2\pi)

. The key difference from sine: cosine is decreasing on

[0, \pi]

and increasing on

[\pi, 2\pi]

, so the shape of the solution interval is different.

$\cos(x) > a$ on $[0, 2\pi)$ : the arc where cosine exceeds $a$ is $[0, x_1) \cup (x_2, 2\pi)$ — the region near $\theta = 0$ (and $2\pi$ ), where the $x$ -coordinate on the unit circle is large.
$\cos(x) < a$ on $[0, 2\pi)$ : the complementary arc $(x_1, x_2)$ .

The boundary cases (

a = \pm 1

|a| > 1

) parallel those for sine.

Tangent inequalities have a different structure because tangent has period

\pi

, vertical asymptotes, and unbounded range. The equation

\tan(x) = a

has one solution per period:

x_0 = \arctan(a)

. Within one period

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

$\tan(x) > a$ : the interval $(x_0, \frac{\pi}{2})$ — from the boundary angle to the next asymptote.
$\tan(x) < a$ : the interval $(-\frac{\pi}{2}, x_0)$ — from the asymptote to the boundary angle.

Since tangent is strictly increasing on each period interval, there is no ambiguity: the function is below

a

to the left of

x_0

and above

a

to the right. The general solution adds

n\pi

to shift across periods.

The asymptotes must always be excluded from the solution, regardless of whether the inequality is strict or non-strict — tangent is undefined there.

Compound Inequalities

A compound trigonometric inequality imposes two simultaneous bounds. For example:

\frac{1}{2} < \sin(x) \leq \frac{\sqrt{3}}{2}

This requires

\sin(x) > \frac{1}{2}

and

\sin(x) \leq \frac{\sqrt{3}}{2}

simultaneously.

Step 1: Solve each inequality separately.

\sin(x) > \frac{1}{2}

[0, 2\pi)

x \in \left(\frac{\pi}{6}, \frac{5\pi}{6}\right)

\sin(x) \leq \frac{\sqrt{3}}{2}

[0, 2\pi)

x \in \left[0, \frac{\pi}{3}\right] \cup \left[\frac{2\pi}{3}, 2\pi\right)

Step 2: Intersect the solution sets.

The intersection is the set of

x

values that satisfy both conditions:

x \in \left(\frac{\pi}{6}, \frac{\pi}{3}\right] \cup \left[\frac{2\pi}{3}, \frac{5\pi}{6}\right)

Graphically, this corresponds to the region where the sine curve lies strictly above

y = \frac{1}{2}

and at or below

y = \frac{\sqrt{3}}{2}

— the band between the two horizontal lines, with appropriate endpoint inclusion.

The unit circle interpretation is equally clear: identify the arcs satisfying each condition, then take their overlap. For the example above, the first condition selects the upper arc from

\frac{\pi}{6}

\frac{5\pi}{6}

, and the second condition excludes the portion between

\frac{\pi}{3}

and

\frac{2\pi}{3}

(where sine exceeds

\frac{\sqrt{3}}{2}

). What remains is two symmetric segments flanking the excluded zone.

Inequalities Involving Transformed Functions

When the trigonometric function is transformed —

2\sin(3x) - 1 > 0

, for instance — the inequality must be handled in stages.

Step 1: Isolate the trigonometric expression. Rearrange to place the function on one side:

2\sin(3x) > 1 \quad \Rightarrow \quad \sin(3x) > \frac{1}{2}

Step 2: Solve for the inner argument. Let

u = 3x

. The inequality

\sin(u) > \frac{1}{2}

has the solution:

u \in \left(\frac{\pi}{6} + 2n\pi, \quad \frac{5\pi}{6} + 2n\pi\right)

Step 3: Recover $x$. Divide by 3:

x \in \left(\frac{\pi}{18} + \frac{2n\pi}{3}, \quad \frac{5\pi}{18} + \frac{2n\pi}{3}\right)

Step 4: Restrict to the target interval. On

[0, 2\pi)

, substitute

n = 0, 1, 2

(since the period of

\sin(3x)

\frac{2\pi}{3}

, three periods fit within

[0, 2\pi)

$n = 0$ : $\left(\frac{\pi}{18}, \frac{5\pi}{18}\right)$
$n = 1$ : $\left(\frac{\pi}{18} + \frac{2\pi}{3}, \frac{5\pi}{18} + \frac{2\pi}{3}\right) = \left(\frac{13\pi}{18}, \frac{17\pi}{18}\right)$
$n = 2$ : $\left(\frac{\pi}{18} + \frac{4\pi}{3}, \frac{5\pi}{18} + \frac{4\pi}{3}\right) = \left(\frac{25\pi}{18}, \frac{29\pi}{18}\right)$

The multiplier in the argument multiplies the number of solution intervals within a fixed domain — the same phenomenon seen in multiple-angle equations. A coefficient of

B

\sin(Bx)

creates

B

copies of the basic solution pattern within one

2\pi

interval.

Vertical shifts (

+D

) and amplitude changes (

A

) affect the threshold value after isolation. Phase shifts (

-C

) offset the interval boundaries. But the core method remains the same: isolate, solve for the inner argument, recover the outer variable, and restrict to the domain.

Domain Considerations for Reciprocal and Quotient Functions

Inequalities involving tangent, cotangent, secant, or cosecant require attention to the domains of these functions — specifically, the vertical asymptotes where they are undefined.

Example: Solve

\sec(x) \geq 2

[0, 2\pi)

.

Rewrite using the reciprocal:

\frac{1}{\cos(x)} \geq 2

. This requires two conditions simultaneously:

1.

\cos(x) > 0

(secant is positive only when cosine is positive)
2.

\cos(x) \leq \frac{1}{2}

(taking the reciprocal of both sides reverses the inequality, since

\cos(x) > 0

)

The intersection of

\cos(x) > 0

and

\cos(x) \leq \frac{1}{2}

[0, 2\pi)

\cos(x) > 0

\left[0, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right)

\cos(x) \leq \frac{1}{2}

\left[\frac{\pi}{3}, \frac{5\pi}{3}\right]

Intersection:

\left[\frac{\pi}{3}, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{2}, \frac{5\pi}{3}\right]

The points

x = \frac{\pi}{2}

and

x = \frac{3\pi}{2}

are excluded because secant is undefined there (vertical asymptotes on the graph).

The reversal of the inequality direction when taking the reciprocal deserves emphasis. If

\cos(x) > 0

and

\frac{1}{\cos(x)} \geq 2

, then

\cos(x) \leq \frac{1}{2}

. If

\cos(x) < 0

and

\frac{1}{\cos(x)} \geq 2

, there is no solution — a negative reciprocal cannot exceed 2. This case analysis — splitting based on the sign of the denominator — is essential for all inequalities involving reciprocal or quotient functions.

Summary of Inequality Forms and Their Solution Approaches

Trigonometric inequalities split into a handful of recognizable forms, each with its own first move and characteristic solution structure. The table below maps every form discussed above to the starting step, the shape of its solution set, and the period over which the pattern repeats.

Inequality form	First step	Solution structure	Repeats with period
Basic: f(x) ⋛ a, \|a\| < 1	solve f(x) = a for the boundary angles	union of intervals; open or closed by strictness	2π (sin, cos, sec, csc); π (tan, cot)
Range-edge: f(x) ⋛ ±1 for sin or cos	compare a to the function's range [−1, 1]	isolated points (at the extrema), all reals, or empty	2π if isolated; n/a if all reals or empty
Out-of-range: \|a\| > 1 for sin or cos	check whether a lies outside [−1, 1] — no boundary equation needed	all reals or empty	n/a
Compound: a ⋛ f(x) ⋛ b	solve each one-sided inequality separately	intersection of the two solution sets	period of f
Transformed: A·f(Bx+C)+D ⋛ a	isolate f(·); substitute u = Bx+C and solve in u	basic-form solution in u, scaled by 1/\|B\| and shifted; \|B\| copies per outer period	period of f scaled by 1/\|B\|
Reciprocal: 1/f(x) ⋛ a (sec, csc)	split by sign of f(x); take the reciprocal carefully on each branch	intersection of sign condition and reciprocal-form inequality; asymptotes always excluded	period of f