Unit 3: Trigonometric & Polar Functions

Trigonometric Equations & Inequalities

Solving trigonometric equations and inequalities using inverse trig functions.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

Sinusoidal Modeling

Finding Every Solution, Not Just the One a Calculator Gives

A sinusoidal model answers 'what output does this input produce?' Before reading on: if you're instead given an output and asked for every input that produces it, why might a single inverse-trig button press on a calculator not be enough?

Inverse trig functions like arcsin only return one value — the principal value, inside a restricted range — because sine and cosine repeat their outputs infinitely often and aren't one-to-one over all real numbers. Solving a trig equation genuinely means finding every angle that works, which means combining the principal value with the function's own symmetry and periodicity, not just reading off one number.

Worked Example — Solving a Basic Trig Equation

Solve 2sin(x) − 1 = 0 for x in [0, 2π). Isolate: sin(x) = 1/2. arcsin(1/2) = π/6 gives the principal solution. Sine is also positive in Quadrant II, at the reflection π − π/6 = 5π/6 (check: sin(5π/6) = sin(π/6) = 1/2, confirmed). Both lie in [0, 2π), so the solutions are x = π/6 and x = 5π/6.

Worked Example — Solving a Trig Inequality

Solve sin(x) ≥ 1/2 on [0, 2π). The boundary values, from the equation above, are x = π/6 and x = 5π/6. Sine reaches its maximum of 1 at x = π/2, which lies between those two boundaries, so sine is ≥ 1/2 on the closed interval between them: [π/6, 5π/6]. Check a point inside, x = π/2: sin(π/2) = 1 ≥ 1/2 — confirmed. Check a point outside, x = 0: sin(0) = 0, correctly excluded.

Worked Example — Solving a Quadratic-Form Trig Equation

Solve 2cos²(x) − 1 = 0 on [0, 2π). This is quadratic in cos(x): cos²(x) = 1/2, so cos(x) = ±√(1/2) = ±√2/2. Each sign gives two solutions from the unit circle: cos(x) = √2/2 at x = π/4 and x = 7π/4; cos(x) = −√2/2 at x = 3π/4 and x = 5π/4. Four solutions total: π/4, 3π/4, 5π/4, 7π/4 — one in each quadrant, matching where cosine equals ±√2/2.

Graph Visualizer

Trace modePoint mode (click graph to plot)

Show x^2

Color

Expression

Domain & range

x min

x max

y min

y max

Precision (decimal places)2

Evaluate a point

Evaluate at x =

x^2 = 0

Tip

Sketch (or graph) the horizontal line at the target output value alongside the trig function first — every intersection point is a solution, and seeing them at once prevents stopping after only the principal value.

Common Mistakes

Reporting only the principal value from arcsin or arccos as if it were the complete solution set.
arcsin and arccos each return exactly one number by definition — a full solution set almost always needs a second family of solutions from the function's own symmetry, plus every repeat from periodicity if the interval isn't restricted.
Taking a square root in a quadratic-form equation like cos²(x) = 1/2 and keeping only the positive root.
Taking a square root of both sides of an equation produces both a positive and a negative case — cos(x) = √2/2 and cos(x) = −√2/2 are both valid branches, each with its own solutions.

Key Takeaways

Inverse trig functions return only a principal value; a complete solution set requires adding the function's other same-output family from symmetry.
A trig inequality's solution set is the interval between the boundary angles found by solving the corresponding equation.
An equation that's quadratic in sin(x) or cos(x) produces two separate cases (from ±) to solve, each contributing its own solutions.

Summary

Solving trig equations depends on reading angles off the unit circle in rectangular (x, y) terms. The next lesson introduces a coordinate system built around angle and distance directly — polar coordinates — the natural language for curves that a rectangular equation struggles to describe.

Track your progress

← Sinusoidal Modeling Back to Unit 3: Trigonometric & Polar Functions The Polar Coordinate System→