How many solutions does a trigonometric equation have?

Usually infinitely many, because trigonometric functions are periodic: if x is a solution, so is x plus a full period. That is why the solution is written with a parameter k running over the integers, and why exam questions almost always restrict the answer to a given interval.

How do I solve sin x = c?

First check that |c| ≤ 1 — otherwise there is no solution. Then find the base solution α (from the table of values or with arcsin) and write two families: x = α + 2kπ and x = π − α + 2kπ. The second one comes from sin(π − α) = sin α.

Why does the cosine equation give ±α while the sine gives α and π − α?

Because the cosine is even: cos(−α) = cos α, so its solutions sit symmetrically about zero. The sine is odd but satisfies sin(π − α) = sin α, so its solutions are symmetric about π/2. These are the same reduction formulas, read as a description of the solution set.

Why is it kπ for the tangent rather than 2kπ?

Because the tangent has period π, not 2π — it repeats twice as often as the sine and cosine. So tan x = c has a single family of solutions, x = α + kπ. It is also the only one of the three equations that has a solution for every value of c.

How do I pick the solutions inside a given interval?

Write the general solution, then substitute successive integers k (…, −1, 0, 1, 2, …) and keep the results that land inside the interval. The graph is the check: the number of solutions must equal the number of crossings of the horizontal line over that stretch.

How do I solve an equation with two different functions in it?

Reduce it to one function. Break up a doubled angle with sin 2x = 2 sin x cos x and swap squares using the Pythagorean identity. Then either factor out a common term and use that a product is zero only when one of its factors is, or substitute t = sin x and solve a quadratic.

What if the substitution gives t = 2?

Discard it. The substitution t = sin x carries the condition −1 ≤ t ≤ 1, so a value outside that range corresponds to no angle at all. It is the most commonly missed step in quadratic equations with a sine.

How are trigonometric inequalities solved?

First solve the corresponding equation, then read off the graph which stretches lie above (or below) the horizontal line. For example sin x ≥ 1/2 on [0, 2π) gives the interval from π/6 to 5π/6 — exactly the stretch where the sine curve runs above the line y = 1/2.

Advanced level

Trigonometric equations

The equation sin x = 1/2 has infinitely many solutions, because the sine repeats. See how to find the base solution, how to write all the others with a parameter k, and how to pick out the ones that fall inside a given interval.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Trigonometric graphsA sine wave is the unit circle unrolled along an axis. See where the period of 2π comes from, why the cosine is a shifted sine, and what happens to the graph of the tangent where the function does not exist.
Reduction formulasReduction formulas bring the sine, cosine and tangent of any angle back to an acute one. Instead of memorising a dozen identities, see the single rule behind them: at 180° and 360° the function stays, at 90° and 270° it swaps for its cofunction, and the sign comes from the quadrant.

Where this is used

Real situations where you count exactly the way this lesson teaches:

Tides
The depth in a harbour varies as h(t) = 2.5 + 1.8 · sin(2πt/12.4) metres, with t in hours. A yacht drawing 3.4 m can enter when sin(2πt/12.4) ≥ 0.5, that is when 2πt/12.4 lies between π/6 and 5π/6 — a window from t = 1.03 h to t = 5.17 h, four hours and eight minutes of high water. Harbour authorities publish exactly this number as the "tidal window".
Mains electricity
The mains voltage is u(t) = 325 · sin(100πt). When does it first reach 230 V? Solve sin(100πt) = 230/325 = 0.708, so 100πt = 0.786 rad and t = 0.0025 s. The second solution in the same half-cycle is 100πt = π − 0.786, that is t = 0.0075 s — between them the voltage stays above 230 V for five milliseconds.
A Ferris wheel
A gondola on a wheel of radius 20 m with its hub 22 m up has height h(t) = 22 − 20 · cos(2πt/9), where t is minutes since boarding. It reaches 32 m when cos(2πt/9) = −0.5, that is 2πt/9 = 2π/3, after t = 3 minutes — and again at t = 6 minutes, on the way back down.
Farming and day length
Day length in central Poland is approximated by d(n) = 12.23 + 4.52 · sin(2π(n − 80)/365) hours, where n is the day of the year. Many crops are sown once the day reaches 14 h, that is sin(2π(n − 80)/365) ≥ 0.392. The first solution falls at n ≈ 103, about 13 April, and the last at n ≈ 237, about 25 August.
Camshaft timing
A valve follower rises by s(φ) = 8 · sin φ millimetres, where φ is the camshaft angle. The valve opens at a lift of 4 mm, that is at sin φ = 0.5: the opening angle is φ = 30° and the closing one φ = 150°. The valve therefore stays open through 120° of shaft rotation, which at 3000 rpm lasts 6.7 ms.

All formulas

Equation with a sine
$\sin x = c \iff x = \alpha + 2k\pi \ \lor \ x = \pi - \alpha + 2k\pi$
α is the base solution; k is any integer
Equation with a cosine
$\cos x = c \iff x = \alpha + 2k\pi \ \lor \ x = -\alpha + 2k\pi$
the solutions sit symmetrically about the x axis
Equation with a tangent
$\tan x = c \iff x = \alpha + k\pi$
one family only — the tangent has period π, not 2π
Condition for a solution
$|c| > 1 \implies \sin x = c \ \text{has no solution}$
the same for the cosine; the tangent takes every value
Special cases
$\sin x = 0 \iff x = k\pi$
and cos x = 0 ⟺ x = π/2 + kπ
Reducing to one function
$\sin 2x = 2\sin x \cos x$
the doubled angle broken up before factoring

The equation $2x + 1 = 7$ has one solution. The equation $\sin x = \frac{1}{2}$ has infinitely many — not through some trick, but because the sine repeats. The whole technique comes down to two questions: what is the base solution, and how does it repeat.

The equation seen on the graph

Solving $\sin x = \frac{1}{2}$ means finding the $x$ at which the graph of $y = \sin x$ has height $\frac{1}{2}$ . That is: cutting the sine curve with a horizontal line.

Four crossings on this stretch, and infinitely many outside it — the curve repeats every 2π.

The drawing shows everything that matters:

Over one period there are two crossings, not one.
Every further period repeats exactly those two, shifted by $2\pi$ .
If the line sat above $y = 1$ , there would be no crossings at all.

The equation with a sine

Write that as a formula. Let $\alpha$ be the base solution, the one in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ — read off the table of values or found with arcsin. The second solution in the period comes from the reduction formula $\sin(\pi - \alpha) = \sin \alpha$ :

\sin x = c \iff x = \alpha + 2k\pi \quad \lor \quad x = \pi - \alpha + 2k\pi, \qquad k \in \mathbb{Z}

There is one existence condition: the range of the sine is $[-1, 1]$ , so

|c| > 1 \implies \text{no solutions}

Three values have only one solution per period, because the two families collapse onto it:

\sin x = 0 \iff x = k\pi, \qquad \sin x = 1 \iff x = \frac{\pi}{2} + 2k\pi, \qquad \sin x = -1 \iff x = -\frac{\pi}{2} + 2k\pi

The equation with a cosine

Here the second solution comes from evenness: $\cos(-\alpha) = \cos \alpha$ , so the crossings sit symmetrically about zero.

The solutions of cos x = 1/2 are symmetric about the y axis, which is what the compact form x = ±π/3 + 2kπ records.

\cos x = c \iff x = \alpha + 2k\pi \quad \lor \quad x = -\alpha + 2k\pi, \qquad k \in \mathbb{Z}

or, shorter, $x = \pm\alpha + 2k\pi$ . The existence condition is the same as for the sine.

The equation with a tangent

The tangent has period $\pi$ rather than $2\pi$ , so there is only one family:

\tan x = c \iff x = \alpha + k\pi, \qquad k \in \mathbb{Z}

It is also the only one of the three that has a solution for every value of $c$ — the range of the tangent is the whole real line.

Solutions inside an interval

Exam questions rarely ask for the whole set; they ask for the solutions in a given stretch. The method is always the same: general formula, then successive integers $k$ .

Solve sin x = −√2/2 on the interval [0, 2π).

Reducible equations

Most problems do not arrive in the form $\sin x = c$ . Two techniques handle almost all of them.

Factoring. When a common factor appears, take it outside the bracket and use that a product is zero only when one of its factors is.

Solve sin 2x = sin x on the interval [0, 2π).

Substitution. When one function appears in several powers, substitute a variable for it and solve a quadratic equation.

Solve 2cos²x + cos x − 1 = 0 on the interval [0, 2π).

Inequalities: the same graph, a different answer

A trigonometric inequality is solved in two steps: first the equation, then a reading off the graph of which side of the line the curve is on.

For $\sin x \geqslant \frac{1}{2}$ on $[0, 2\pi)$ the equation gives $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ , and the first drawing of this lesson shows that between those points the sine curve runs above the line. Hence

x \in \left[ \frac{\pi}{6}, \frac{5\pi}{6} \right]

The answer to an inequality is therefore an interval rather than a list of points — which is why the graph here is a tool rather than an illustration.

Exercises

Where one solution from an interval is asked for, type the number of degrees (e.g. 150). Where two solutions in $[0, 2\pi)$ are asked for, give both, in increasing order, as multiples of π — e.g. π/6, 5π/6. Where the number of solutions is asked for, the answer is a whole number.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

sin x = √2/2, x ∈ [0°, 90°] → x

Common mistakes

Giving only one solution per period — the sine and the cosine cross a horizontal line twice in a period; a calculator (and the arcus function) returns only one of them.
Cancelling the equation by $\sin x$ — dividing by an expression that may be zero loses solutions. Factor it out instead.
Writing $2k\pi$ for the tangent — the tangent has period $\pi$ , so its family is $\alpha + k\pi$ .
Skipping the condition $|c| \leqslant 1$ — $\sin x = \frac{3}{2}$ has no solutions, and that is a complete answer rather than a missing one.
Forgetting the condition on a substitution — after $t = \cos x$ , roots outside $[-1, 1]$ have to be discarded.
Mixing degrees and radians in one answer — if the interval is given as $[0, 2\pi)$ , the solutions are written in radians too.

Formula card

Topic: Trigonometric equations

Equation with a sine
$\sin x = c \iff x = \alpha + 2k\pi \ \lor \ x = \pi - \alpha + 2k\pi$
α is the base solution; k is any integer
Equation with a cosine
$\cos x = c \iff x = \alpha + 2k\pi \ \lor \ x = -\alpha + 2k\pi$
the solutions sit symmetrically about the x axis
Equation with a tangent
$\tan x = c \iff x = \alpha + k\pi$
one family only — the tangent has period π, not 2π
Condition for a solution
$|c| > 1 \implies \sin x = c \ \text{has no solution}$
the same for the cosine; the tangent takes every value
Special cases
$\sin x = 0 \iff x = k\pi$
and cos x = 0 ⟺ x = π/2 + kπ
Reducing to one function
$\sin 2x = 2\sin x \cos x$
the doubled angle broken up before factoring

The equation sin x = 1/2 on the graph: the solutions are the x-coordinates where the sine curve crosses the horizontal line y = 1/2. There are infinitely many, because the curve repeats.

The same for the cosine: the solutions of cos x = 1/2 sit symmetrically on either side of zero — hence the compact form x = ±π/3 + 2kπ.

Frequently asked questions

A luminous unit circle with a radius sweeping out an angle, a sine wave unrolling from it as a streak of light, beside a right triangle drawn in stars, against a cosmic nebula in violet and magenta

Trigonometry

Why sine doesn't depend on the size of the triangle

The height of a tower, the width of a river, the distance to a ship offshore — all of it can be computed without going anywhere near the thing being measured. The whole of trigonometry rests on one observation: the ratio of two sides in a right triangle depends only on the angle, not on how big the triangle is. That is what turns an angle into a number — and lets it be tabulated once and for all.

13 min