Trigonometric Equations

A trigonometric equation asks a deceptively simple question: for what angles does a trig function take a given value? Because sine, cosine, and tangent repeat forever, that question almost never has a single answer — it has infinitely many, spaced out by the function's period. Learning to solve trig equations means learning to capture all of those answers in one compact formula while still being able to pick out the specific solutions a problem cares about.

This skill sits at the crossroads of algebra, geometry, and the physics of anything that oscillates. Once you can solve $\sin x = \tfrac{1}{2}$ completely, you can answer questions like "at what times is the tide exactly two metres high?" or "for which angles does this pendulum return to the centre?" — the same mathematical machinery, applied everywhere something repeats.

Learning Objectives

• Solve basic trigonometric equations of the form $\sin x = k$, $\cos x = k$, and $\tan x = k$.
• Write general solutions that capture every angle satisfying an equation, using the integer $n$.
• Understand why the period of each function dictates the spacing of solutions.
• Restrict a general solution to a given interval such as $[0, 2\pi)$.
• Use algebraic techniques (factoring, substitution) and trigonometric identities to solve more complex equations.
• Recognise and avoid the classic errors: lost solutions, out-of-range values, and dividing away roots.

Quick Answer

To solve a trig equation, first isolate the trig function, then find one reference solution, then add the period to generate all others. For sine, the general solution of $\sin x = k$ is $x = \arcsin k + 2\pi n$ or $x = \pi - \arcsin k + 2\pi n$. For cosine, $\cos x = k$ gives $x = \pm \arccos k + 2\pi n$. For tangent, $\tan x = k$ gives $x = \arctan k + \pi n$ (period $\pi$, not $2\pi$). Here $n$ is any integer. More complicated equations are attacked by factoring, substituting a single trig function, or applying identities to make every term share the same function and angle. Always check that proposed values lie in range ($-1 \le \sin, \cos \le 1$) and inside the required interval.

Where It Came From

Trigonometric equations were not born in a classroom — they were forced into existence by the sky. Ancient astronomers needed to predict repeating events: the rising of stars, the phases of the Moon, the recurring positions of the planets. These are inherently periodic problems, and periodic problems demand periodic functions. Hipparchus of Nicaea (around 150 BCE) built the first known table of chords — an ancestor of the sine table — precisely so he could solve for unknown angles in the triangles formed by celestial sightlines. To find when a star would reach a certain altitude, he was, in effect, solving a trigonometric equation.

The tradition deepened in India, where mathematicians such as Aryabhata (around 500 CE) tabulated the sine ("jya") to predict eclipses and planetary conjunctions. Islamic astronomers including Al-Battani refined these methods for calendar and prayer-time calculations, both fundamentally about solving "for which angle (or time) does the Sun reach this position?"

The subject became truly general when Leonhard Euler, in the 1700s, recast trigonometry in terms of functions rather than triangle ratios, and Joseph Fourier (early 1800s) showed that any periodic phenomenon — heat flow, vibrating strings, later electrical signals — could be built from sines and cosines. At that moment solving trig equations stopped being an astronomer's niche tool and became the universal language of oscillation. Every time an engineer today asks "when does this waveform cross zero?", they are answering Hipparchus's question with Euler's functions.

The Core Idea: One Solution Becomes Infinitely Many

Suppose we want every angle $x$ with

$\sin x = \frac{1}{2}.$

Your calculator gives $\arcsin(0.5) = \tfrac{\pi}{6}$ (30°). But the sine of $\tfrac{5\pi}{6}$ (150°) is also $\tfrac{1}{2}$, because sine is positive in both the first and second quadrants and $\sin(\pi - \theta) = \sin \theta$. So within one turn of the circle there are two solutions: $\tfrac{\pi}{6}$ and $\tfrac{5\pi}{6}$.

Now the key: sine has period $2\pi$, so we can add or subtract any whole number of full turns and still land on the same value. The complete general solution is therefore

$x = \frac{\pi}{6} + 2\pi n \quad \text{or} \quad x = \frac{5\pi}{6} + 2\pi n, \qquad n \in \mathbb{Z}.$

Worked example — listing solutions in an interval. Find all $x$ in $[0, 2\pi)$ with $\sin x = \tfrac{1}{2}$. Take $n = 0$ in both branches: $x = \tfrac{\pi}{6}$ and $x = \tfrac{5\pi}{6}$. Trying $n = 1$ pushes us past $2\pi$, and $n = -1$ gives negatives. So the interval contains exactly two solutions: $\tfrac{\pi}{6}$ and $\tfrac{5\pi}{6}$.

General-Solution Formulas and Why Cosine and Tangent Differ

Each function's symmetry determines the shape of its general solution.

Sine ($\sin x = k$, with $-1 \le k \le 1$): $x = \arcsin k + 2\pi n \quad \text{or} \quad x = \pi - \arcsin k + 2\pi n.$ A compact single form is $x = \pi n + (-1)^n \arcsin k$.

Cosine ($\cos x = k$, with $-1 \le k \le 1$): cosine is even, so if $\theta$ works then $-\theta$ works too: $x = \pm \arccos k + 2\pi n.$

Tangent ($\tan x = k$, any real $k$): tangent repeats every $\pi$, not $2\pi$, so its solutions are twice as dense: $x = \arctan k + \pi n.$

Worked example — cosine. Solve $\cos x = -\tfrac{\sqrt{3}}{2}$ on $[0, 2\pi)$. The reference angle is $\arccos\!\left(\tfrac{\sqrt{3}}{2}\right) = \tfrac{\pi}{6}$, and cosine is negative in quadrants II and III. The general solution is $x = \pm \tfrac{5\pi}{6} + 2\pi n$. Taking the $+$ branch with $n=0$ gives $\tfrac{5\pi}{6}$; the $-$ branch with $n=0$ gives $-\tfrac{5\pi}{6}$, which is outside the interval, so add $2\pi$ to get $\tfrac{7\pi}{6}$. Solutions: $\tfrac{5\pi}{6}$ and $\tfrac{7\pi}{6}$.

Worked example — tangent. Solve $\tan x = 1$ on $[0, 2\pi)$. General solution $x = \tfrac{\pi}{4} + \pi n$. With $n = 0$: $\tfrac{\pi}{4}$. With $n = 1$: $\tfrac{5\pi}{4}$. With $n = 2$: $\tfrac{9\pi}{4} > 2\pi$, stop. Solutions: $\tfrac{\pi}{4}$ and $\tfrac{5\pi}{4}$. Notice both fall on the same "line" through the origin — that is tangent's $\pi$-periodicity showing itself.

Equations With a Multiple Angle

When the argument is not just $x$, solve for the whole argument first, then divide — and be careful to generate enough solutions before dividing.

Worked example. Solve $\cos 2x = \tfrac{1}{2}$ on $[0, 2\pi)$. Let $u = 2x$. Since $x \in [0, 2\pi)$, we have $u \in [0, 4\pi)$ — a doubled interval, so expect twice as many solutions. Now $\cos u = \tfrac{1}{2}$ gives $u = \pm \tfrac{\pi}{3} + 2\pi n$. Listing values of $u$ in $[0, 4\pi)$: $\tfrac{\pi}{3}, \tfrac{5\pi}{3}, \tfrac{7\pi}{3}, \tfrac{11\pi}{3}$. Divide each by 2: $x = \frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{7\pi}{6},\ \frac{11\pi}{6}.$ Four solutions, exactly as the doubled interval predicted.

Solving by Factoring and Substitution

Many equations become tractable once you treat a trig function as a single variable.

Worked example — factoring. Solve $2\sin^2 x - \sin x = 0$ on $[0, 2\pi)$. Factor: $\sin x\,(2\sin x - 1) = 0.$ So $\sin x = 0$ or $\sin x = \tfrac{1}{2}$. From $\sin x = 0$: $x = 0, \pi$. From $\sin x = \tfrac{1}{2}$: $x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$. All four: $0,\ \tfrac{\pi}{6},\ \tfrac{5\pi}{6},\ \pi$. (Crucially, we did not divide both sides by $\sin x$ — doing so would have thrown away the $\sin x = 0$ roots.)

Using Identities to Solve

If an equation mixes different functions or angles, use identities to force everything into terms of one function.

Worked example. Solve $2\cos^2 x + \sin x = 1$ on $[0, 2\pi)$. Replace $\cos^2 x$ with $1 - \sin^2 x$ using the Pythagorean identity: $2(1 - \sin^2 x) + \sin x = 1.$ Expand and rearrange: $2 - 2\sin^2 x + \sin x = 1 \;\Longrightarrow\; 2\sin^2 x - \sin x - 1 = 0.$ Let $s = \sin x$: $2s^2 - s - 1 = 0$, which factors as $(2s + 1)(s - 1) = 0$, giving $s = -\tfrac{1}{2}$ or $s = 1$.

• $\sin x = 1 \Rightarrow x = \tfrac{\pi}{2}$.
• $\sin x = -\tfrac{1}{2} \Rightarrow x = \tfrac{7\pi}{6}, \tfrac{11\pi}{6}$.

Solutions on $[0, 2\pi)$: $\tfrac{\pi}{2},\ \tfrac{7\pi}{6},\ \tfrac{11\pi}{6}$.

Real-World Applications

• Tides and daily cycles. Tide height is modelled as $h(t) = A\cos(\omega t) + d$. Asking "when is the water 2 m deep?" is solving a cosine equation for $t$ — vital for shipping and harbour navigation.
• Alternating current. A voltage $V = V_0 \sin(\omega t)$ hits a safety threshold at times found by solving $\sin(\omega t) = V/V_0$; engineers use these zero- and level-crossings to time circuits.
• Astronomy and navigation. Determining when the Sun reaches a given altitude (for sunrise tables, solar-panel angles, or celestial navigation) reduces to a trig equation in the hour angle.
• Sound and music. Finding when two tones interfere constructively means solving where their sine waves align — the mathematics behind beats and tuning.
• Circular and orbital motion. Locating when a rotating point returns to a specific height or phase is a direct trig-equation problem.

Common Mistakes

1. Reporting only the calculator's answer. A student solves $\sin x = \tfrac{1}{2}$ and writes $x = \tfrac{\pi}{6}$, stopping there. Why wrong: the calculator gives one value from the arcsine's restricted range; it hides the second-quadrant solution and all periodic repeats. Correction: always find the sibling solution in the other relevant quadrant and add $2\pi n$ (or $\pi n$ for tangent).

2. Dividing by a trig function. Solving $\sin x \cos x = \sin x$ by dividing both sides by $\sin x$ to get $\cos x = 1$. Why wrong: you silently assumed $\sin x \ne 0$ and deleted every solution where $\sin x = 0$. Correction: move everything to one side and factor: $\sin x(\cos x - 1) = 0$, then solve each factor.

3. Forgetting the interval expands with the angle. For $\sin 3x = 0$ on $[0, 2\pi)$, solving as if the argument ranged over $[0, 2\pi)$ and getting too few answers. Why wrong: if $x \in [0, 2\pi)$ then $3x \in [0, 6\pi)$, so there are three times as many solutions. Correction: solve for the full argument over its stretched interval, then divide.

4. Accepting out-of-range values. Concluding from $2\cos x = 3$ that $\cos x = 1.5$ and hunting for an angle. Why wrong: cosine never exceeds 1, so there is no solution. Correction: check $-1 \le k \le 1$ before solving.

Comparison and Connections

Trig equations are the inverse task of evaluating trig functions, and they lean on identities as tools. The table contrasts the three basic functions:

Function	Range of $k$	Period	General solution of $f(x)=k$	Solutions per $2\pi$
$\sin x = k$	$-1 \le k \le 1$	$2\pi$	$\arcsin k + 2\pi n$ or $\pi - \arcsin k + 2\pi n$	2 (1 at extremes)
$\cos x = k$	$-1 \le k \le 1$	$2\pi$	$\pm \arccos k + 2\pi n$	2 (1 at extremes)
$\tan x = k$	any real	$\pi$	$\arctan k + \pi n$	2 (over $[0,2\pi)$)

Compare this with a polynomial equation, which has finitely many roots: the defining feature of a trig equation is its infinite, periodic solution set. And unlike an identity (true for all $x$, e.g. $\sin^2 x + \cos^2 x = 1$), an equation is only true for particular $x$ — identities are the tools, equations are the puzzles.

Practice Questions

Recall

State the general solution of $\tan x = \sqrt{3}$.

Answer: $\tan x = \sqrt{3}$ has reference angle $\tfrac{\pi}{3}$, so $x = \tfrac{\pi}{3} + \pi n$, $n \in \mathbb{Z}$.

Understanding

Explain why $\cos x = k$ uses $\pm \arccos k$ while $\sin x = k$ uses $\arcsin k$ and $\pi - \arcsin k$.

Guidance: cosine is an even function symmetric about the $x$-axis, so $\theta$ and $-\theta$ share a cosine value. Sine is symmetric about the vertical line $x = \tfrac{\pi}{2}$, so $\theta$ and $\pi - \theta$ share a sine value. The formulas simply encode each function's symmetry.

Application

Solve $2\sin x + 1 = 0$ on $[0, 2\pi)$.

Answer: $\sin x = -\tfrac{1}{2}$. Sine is negative in quadrants III and IV; reference angle $\tfrac{\pi}{6}$. So $x = \pi + \tfrac{\pi}{6} = \tfrac{7\pi}{6}$ and $x = 2\pi - \tfrac{\pi}{6} = \tfrac{11\pi}{6}$.

Analysis

Solve $\tan^2 x - 1 = 0$ on $[0, 2\pi)$ and explain how many solutions you expect before computing.

Answer: Factor as $(\tan x - 1)(\tan x + 1) = 0$, so $\tan x = 1$ or $\tan x = -1$. Because tangent has period $\pi$, each value yields two solutions in $[0, 2\pi)$, so expect four. From $\tan x = 1$: $\tfrac{\pi}{4}, \tfrac{5\pi}{4}$. From $\tan x = -1$: $\tfrac{3\pi}{4}, \tfrac{7\pi}{4}$. Four solutions, as predicted.

FAQ

Why does $n$ appear in every general solution? Because trig functions are periodic. Adding a whole number of periods returns the function to the same value, so every solution has infinitely many "clones" spaced one period apart. The integer $n$ indexes all of them at once.

When do I add $2\pi n$ versus $\pi n$? Use $2\pi n$ for sine and cosine (period $2\pi$) and $\pi n$ for tangent and cotangent (period $\pi$). Match the added term to the function's own period, not the problem's.

How do I know which quadrants to use? Look at the sign of $k$. Sine is positive in quadrants I and II; cosine positive in I and IV; tangent positive in I and III. The reference angle from your calculator locates one solution, and the sign tells you where its partner lives.

What if the equation has no solution? That happens when a required value falls outside a function's range — for example $\sin x = 2$ or $\cos x = -1.4$. Always check $-1 \le k \le 1$ for sine and cosine before solving. Tangent, having unbounded range, always has solutions.

Can I just use a graph instead of formulas? Graphing is an excellent check: the solutions are where the curve $y = f(x)$ meets the horizontal line $y = k$, and you can literally count them in an interval. But graphs give approximate values, so use them to confirm the count and catch missing solutions, then report exact answers from the algebra.

Why did factoring beat dividing in the $\sin x \cos x = \sin x$ example? Dividing by $\sin x$ assumes it is nonzero and erases the $\sin x = 0$ solutions. Factoring keeps every possibility on the table, so no roots are lost. As a rule, never divide an equation by an expression that could be zero.

Quick Revision

• $\sin x = k$: $\;x = \arcsin k + 2\pi n$ or $x = \pi - \arcsin k + 2\pi n$.
• $\cos x = k$: $\;x = \pm \arccos k + 2\pi n$.
• $\tan x = k$: $\;x = \arctan k + \pi n$.
• Periods: sine and cosine $2\pi$; tangent $\pi$.
• Sine and cosine need $-1 \le k \le 1$; tangent takes any real value.
• Multiple angle: solve for the whole argument over its stretched interval, then divide.
• Factor rather than divide, so you never lose roots.
• Use the Pythagorean identity to convert everything into one function, then treat it as an algebraic (often quadratic) equation.
• Always restrict the general solution to the required interval at the end.

Related Topics

Prerequisites

• Trigonometry Overview
• Trigonometric functions and the unit circle
• Inverse trigonometric functions (arcsin, arccos, arctan)

Related Topics

• Trigonometric identities (Pythagorean, double-angle, sum-and-difference)
• Graphs of trigonometric functions and period
• Algebra — quadratic equations and factoring

Next Topics

• Applications to simple harmonic motion and waves
• Calculus — derivatives and integrals of trig functions
• Fourier analysis of periodic signals