Trigonometric Inequality Solver

Math Algebra • Inequalities

Written by STEM Calculators Team Published December 12, 2025 Updated February 24, 2026

Enter a trigonometric inequality and press Calculate.

Unit circle: green arcs (for \(\sin,\cos\)) show angles \(y\) (mod \(2\pi\)) where the inequality holds.

Graph of \(y=f(kx+\varphi)\) (blue) and \(y=c\) (orange dashed). Green segments on the \(x\)-axis show where the inequality holds.

Number line: thick green segments mark the solution intervals. Closed/open dots reflect \(\le,\ge\) vs <,>.

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7. Trigonometric Inequality Solver — Theory

We study inequalities of the form \(f(kx+\varphi)\ \square\ c\) with \(f\in\{\sin,\cos,\tan\}\). The solver uses unit-circle geometry or monotonicity on canonical intervals to obtain “base arcs” in one period, then maps them back to \(x\) via \(y=kx+\varphi\) and intersects with the chosen domain.

1) Background and geometry

The trigonometric functions are periodic in the angle \(y\) measured in radians:

\(\sin(y+2\pi)=\sin y\), \(\cos(y+2\pi)=\cos y\), \(\tan(y+\pi)=\tan y\).
On the unit circle, \(\sin y\) is the \(y\)-coordinate and \(\cos y\) is the \(x\)-coordinate of the point reached by rotating angle \(y\) counter-clockwise from \((1,0)\). The value \(\tan y\) is the slope \(\sin y/\cos y\) of the terminal line (undefined at \(\frac{\pi}{2}+\pi n\)).

Function	Range	Period \(P_y\)	Principal inverse	Monotone on
\(\sin y\)	\([-1,1]\)	\(2\pi\)	\(\arcsin:[-1,1]\to[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\)	Increasing on \([-\tfrac{\pi}{2},\tfrac{\pi}{2}]\)
\(\cos y\)	\([-1,1]\)	\(2\pi\)	\(\arccos:[-1,1]\to[0,\pi]\)	Decreasing on \([0,\pi]\)
\(\tan y\)	\(\mathbb{R}\)	\(\pi\)	\(\arctan:\mathbb{R}\to(-\tfrac{\pi}{2},\tfrac{\pi}{2})\)	Increasing on \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\)

2) Why the “base arcs” look the way they do

Write \(y=kx+\varphi\). First solve \(f(y)\ \square\ c\) for \(y\) inside one period \(P_y\) to obtain base arcs; then repeat every \(P_y\) and map back to \(x\).

\(\sin y \ge c\). Let \(\alpha=\arcsin(c)\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\). By symmetry \(\sin(\pi-\alpha)=\sin\alpha=c\). On the circle the region where \(\sin y\ge c\) is the horizontal belt between \(y=\alpha\) and \(y=\pi-\alpha\). Hence, in one period, \[ y\in[\alpha,\ \pi-\alpha]\quad(\text{endpoints included for } \ge). \] For \(\sin y \le c\), take the complement in \([0,2\pi)\).
\(\cos y \le c\). Let \(\beta=\arccos(c)\in[0,\pi]\). The set where \(\cos y\le c\) is the vertical belt between the rays at angles \(y=\beta\) and \(y=2\pi-\beta\): \[ y\in[\beta,\ 2\pi-\beta]. \] For \(\cos y \ge c\), take the complement.
\(\tan y \ge c\). With \(\gamma=\arctan(c)\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})\), on the monotone branch \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\) we have \[ y\in[\gamma,\ \tfrac{\pi}{2}) \quad (\text{repeat every } \pi). \] Replace \(\ge\) with \(\le\) to flip the inequality interval accordingly.

Boundary cases: if \(|c|>1\) for \(\sin,\cos\), the inequality may be always true or impossible. For \(\tan\) any real \(c\) is admissible, but vertical asymptotes at \(y=\frac{\pi}{2}+\pi n\) split the period and produce open endpoints near the asymptotes.

3) Mapping back to \(x\) and endpoint rules

After we have arcs for \(y\), convert them to \(x\) via \(x=(y-\varphi)/k\).

Period in \(x\): \(P_x=\dfrac{P_y}{|k|}\). If \(k=0\) the left side is constant (degenerate case).
Orientation: if \(k\lt 0\) the interval direction reverses; left/right endpoints swap.
Strict vs non-strict: \(\le,\ge\) include the boundary; \<, \> exclude it. Endpoints at asymptotes are always open.
Domain intersection: finally intersect with \([x_{\min},x_{\max}]\) according to the checkboxes (include/exclude boundaries).

4) Algorithm used by the solver (summary)

Set \(y=kx+\varphi\) and determine \(P_y\) (\(2\pi\) for \(\sin,\cos\); \(\pi\) for \(\tan\)).
Compute base arcs for \(y\) in one period using the rules above; for strict inequalities use open endpoints.
Tile those arcs across \(\mathbb{R}\) by adding integer multiples of \(P_y\).
Map every arc to \(x\) by \(x=(y-\varphi)/k\) (flip if \(k<0\)).
Intersect with the requested domain and merge touching/overlapping pieces.

5) Compact worked examples

Example A. Solve \(\sin(3x)\ge \sqrt{3}/2\) on \([0,2\pi)\).

Base (in \(y\)): \(\alpha=\arcsin(\sqrt{3}/2)=\pi/3\), so \(y\in[\pi/3,\,2\pi/3]\) (mod \(2\pi\)). With \(y=3x\), we get \(x\in[\tfrac{\pi/3}{3},\,\tfrac{2\pi/3}{3}]=[\pi/9,\,2\pi/9]\) (mod \(2\pi/3\)). Tiling by \(P_x=\tfrac{2\pi}{3}\) inside \([0,2\pi)\) yields three blocks:

\[ x\in\Bigl[\tfrac{\pi}{9},\tfrac{2\pi}{9}\Bigr]\ \cup\ \Bigl[\tfrac{\pi}{9}+\tfrac{2\pi}{3},\tfrac{2\pi}{9}+\tfrac{2\pi}{3}\Bigr]\ \cup\ \Bigl[\tfrac{\pi}{9}+\tfrac{4\pi}{3},\tfrac{2\pi}{9}+\tfrac{4\pi}{3}\Bigr]. \]

Example B. Solve \(\cos(2x)\le -\tfrac{\sqrt{2}}{2}\) on \([-\pi,\pi]\).

Base (in \(y\)): \(\beta=\arccos(-\tfrac{\sqrt{2}}{2})=\tfrac{3\pi}{4}\). For “\(\le\)” we keep \(y\in[\beta,\,2\pi-\beta]=[\tfrac{3\pi}{4},\,\tfrac{5\pi}{4}]\) (mod \(2\pi\)). With \(y=2x\) (so \(P_x=\pi\)), mapping gives \(x\in\bigl[\tfrac{3\pi}{8},\,\tfrac{5\pi}{8}\bigr]\) (mod \(\pi\)). Intersecting with \([-\pi,\pi]\) yields two symmetric intervals.

6) Common pitfalls

Degrees vs radians: the solver expects radians (e.g., pi/6, 2*pi/3).
Negative \(k\): remember to reverse interval orientation after mapping from \(y\) to \(x\).
Cosine complements: for \(\cos y \ge c\), do outside of \([\arccos c,\,2\pi-\arccos c]\); for \(\le\) take the inside.
Asymptotes for \(\tan\): endpoints at \(y=\tfrac{\pi}{2}+\pi n\) are always open; strictness applies only to non-asymptotic boundaries.

7) Handy exact values

\(y\)	\(\sin y\)	\(\cos y\)	\(\tan y\)
\(0\)	0	1	0
\(\pi/6\)	\(1/2\)	\(\sqrt{3}/2\)	\(1/\sqrt{3}\)
\(\pi/4\)	\(\sqrt{2}/2\)	\(\sqrt{2}/2\)	1
\(\pi/3\)	\(\sqrt{3}/2\)	\(1/2\)	\(\sqrt{3}\)
\(\pi/2\)	1	0	—

Frequently Asked Questions

How do I solve a trigonometric inequality like sin(kx + phi) >= c?

Convert the argument to y = kx + phi, solve the inequality for y over one period, repeat it by periodicity, and map back to x. Then intersect with your chosen interval [x_min, x_max].

Why does the calculator require radians instead of degrees?

Trigonometric periods and inverse-trig reference angles are naturally defined in radians (2*pi for sin and cos, pi for tan). Using radians keeps the solution intervals and periodic tiling consistent.

What do k and phi mean in sin(kx + phi) or cos(kx + phi)?

k controls the frequency (it scales the period in x), and phi is a horizontal phase shift in the argument. The substitution y = kx + phi is used to map base solution arcs back to x.

How are endpoints handled for < vs ≤ (and > vs ≥)?

Strict inequalities (<, >) exclude boundary points where the expression equals c, while non-strict inequalities (≤, ≥) include them. Domain endpoint checkboxes also control whether x_min and x_max are included.

Why do tangent inequalities have breaks in the solution?

tan(y) has vertical asymptotes at y = pi/2 + pi*n, so solutions are split across monotonic branches. Endpoints at asymptotes are always open even if the inequality is non-strict.

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