Solving Trigonometric Equations

Type an equation and an interval. The page finds the calculator’s principal value, the other value in the same turn (for sine and cosine) and every solution the interval holds, and shows each one twice: where the curve meets the level line, and where the argument’s spiral crosses a solution ray on the circle.

sin x = k, \|k\| ≤ 1	$x = α + 2 nπ, x = π - α + 2 nπ α = arcsin k$
cos x = k, \|k\| ≤ 1	$x = \pm α + 2 nπ, α = arccos k$
tan x = k, any k	$x = α + nπ, α = arctan k$

Worked example — Solving Trigonometric Equations

Principal value, the calculator’s	17.46°
The other value in the same turn	162.54°
General solution, n any integer	162.54° + 360°n
Solutions in the interval	17.46°, 162.54°
Number of solutions	2

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Solving Trigonometric Equations

A trigonometric equation asks for every angle in a given interval at which a trigonometric expression takes a given value. Its general solution drops the interval and lists every angle that solves it, as families x = x₀ + n × period with n any integer: one or two for each value the function must take.

Where the formula comes from

Find the principal value. α = arcsin k is the one value a calculator returns, from a fixed range. It may lie outside the interval and still be where the solving starts.
$arcsin k arccos k arctan k \in [- 9 0^{\circ}, 9 0^{\circ}] \in [0^{\circ}, 18 0^{\circ}] \in (- 9 0^{\circ}, 9 0^{\circ})$
Find the other value in the same turn from the circle’s symmetry: the line y = k meets the circle at α and 180° − α, the line x = k at α and −α, and the line through O of slope k at α and α + 180°.
$sin x = k : cos x = k : tan x = k : α, 18 0^{\circ} - α α, - α α, α + 18 0^{\circ}$
Add whole periods until the argument has covered its interval, then undo the argument. For f(ax + b), u = ax + b runs from a·x₁ + b to a·x₂ + b, and each value of u gives x = (u − b)/a.
$u = u_{0} + 36 0^{\circ} n (tan : u_{0} + 18 0^{\circ} n), x = \frac{u - b}{a}$
Two functions, or a square: make it one function, then a quadratic. Replace cos²x by 1 − sin²x, solve for s = sin x, and keep only the roots between −1 and 1.
$cos^{2} x = 1 - sin^{2} x, s = sin x, - 1 \leq s \leq 1$

Assumes: sin x = k and cos x = k have solutions only for −1 ≤ k ≤ 1; tan x = k has them for every k.

Problems solved in full

sin x = k, in degrees 3 steps
Solve sin x = 0.3 for 0° ≤ x ≤ 360°, to two decimal places.

Open this problem in the tool
1. The principal value is the calculator’s:
  
  $α = arcsin 0.3 = 17.4 6^{\circ}$
2. Sine is positive in quadrants I and II, and the other value in the turn is 180° − α:
  
  $x = 18 0^{\circ} - 17.4 6^{\circ} = 162.5 4^{\circ}$
3. Adding or subtracting 360° takes either out of the interval, so there are no more:
  
  $17.4 6^{\circ} + 36 0^{\circ} = 377.4 6^{\circ} > 36 0^{\circ}, 162.5 4^{\circ} - 36 0^{\circ} = - 197.4 6^{\circ} < 0^{\circ}$
Answer

x = 17.46° or x = 162.54°.

$x = 17.4 6^{\circ}, x = 162.5 4^{\circ}$
cos x = k, on an interval centred at 0 3 steps
Solve cos x = −0.6 for −180° ≤ x ≤ 180°.

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Show solution
1. arccos returns a value from 0° to 180°, here in quadrant II, where cosine is negative:
  
  $α = arccos (- 0.6) = 126.8 7^{\circ}$
2. The other value is −α, the mirror image in the x-axis:
  
  $x = - α = - 126.8 7^{\circ}$
3. Both lie in the interval, and one more period takes either out of it:
  
  $126.8 7^{\circ} - 36 0^{\circ} = - 233.1 3^{\circ} < - 18 0^{\circ}, - 126.8 7^{\circ} + 36 0^{\circ} = 233.1 3^{\circ} > 18 0^{\circ}$
Answer

x = ±126.87°.

$x = \pm 126.8 7^{\circ}$
tan x = k, when the calculator’s value is outside the interval 2 steps
Solve tan x = −2 for 0° ≤ x ≤ 360°.

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Show solution
1. The calculator gives a negative angle, outside the interval:
  
  $α = arctan (- 2) = - 63.4 3^{\circ}$
2. Tangent repeats every 180°, so add 180° once and then twice; a third time gives 476.57°, past 360°:
  
  $x = - 63.4 3^{\circ} + 18 0^{\circ} = 116.5 7^{\circ}, x = - 63.4 3^{\circ} + 36 0^{\circ} = 296.5 7^{\circ}$
Answer

x = 116.57° or x = 296.57°. The calculator’s −63.43° is neither.

$x = 116.5 7^{\circ}, x = 296.5 7^{\circ}$
An exact answer in radians 3 steps
Solve sin x = −√3/2 for 0 ≤ x ≤ 2π, giving exact answers.

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Show solution
1. Since sin(π/3) = √3/2, the principal value is its negative:
  
  $α = arcsin (- \frac{3}{2}) = - \frac{π}{3}$
2. The other value in the turn is π − α:
  
  $x = π - (- \frac{π}{3}) = \frac{4 π}{3}$
3. α itself is negative, so add 2π to bring it into the interval:
  
  $x = - \frac{π}{3} + 2 π = \frac{5 π}{3}$
Answer

x = 4π/3 or x = 5π/3.

$x = \frac{4 π}{3}, x = \frac{5 π}{3}$
A compound argument changes the count 3 steps
Solve sin(2x − 30°) = √3/2 for 0° ≤ x ≤ 360°.

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Show solution
1. Let u = 2x − 30°. As x runs from 0° to 360°, u runs over two full turns:
  
  $u = 2 x - 3 0^{\circ} : 2 \cdot 0^{\circ} - 3 0^{\circ} = - 3 0^{\circ} \leq u \leq 2 \cdot 36 0^{\circ} - 3 0^{\circ} = 69 0^{\circ}$
2. In one turn sin u = √3/2 at 60° and 120°; one more turn adds 360° to each:
  
  $u = 6 0^{\circ}, 12 0^{\circ}, 42 0^{\circ}, 48 0^{\circ}$
3. Undo the argument, x = (u + 30°)/2:
  
  $x = \frac{u + 3 0 ^{\circ}}{2} = 4 5^{\circ}, 7 5^{\circ}, 22 5^{\circ}, 25 5^{\circ}$
Answer

Four solutions: x = 45°, 75°, 225° and 255°. Solving for u only from 0° to 360° would have found two of them.

$x = 4 5^{\circ}, 7 5^{\circ}, 22 5^{\circ}, 25 5^{\circ}$
A quadratic in cos x with a root to reject 3 steps
Solve 2cos²x + 3cos x − 2 = 0 for 0° ≤ x ≤ 360°.

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Show solution
1. Put s = cos x and factorise:
  
  $2 s^{2} + 3 s - 2 = (2 s - 1) (s + 2) = 0$
2. s = −2 is a root of the quadratic but not a value cosine can take, so only one equation is left:
  
  $s = \frac{1}{2}, s = - 2 < - 1$
3. cos x = 1/2 gives α = 60°, and the other value −60° becomes 300° in the interval:
  
  $x = 6 0^{\circ}, x = 36 0^{\circ} - 6 0^{\circ} = 30 0^{\circ}$
Answer

x = 60° or x = 300°.

$x = 6 0^{\circ}, x = 30 0^{\circ}$
Using sin²x + cos²x = 1 to get one function 3 steps
Solve 2cos²x + 3sin x − 3 = 0 for 0° ≤ x ≤ 360°.

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Show solution
1. The equation has a cosine and a sine. Replace cos²x by 1 − sin²x, so that only sin x is left:
  
  $2 (1 - sin^{2} x) + 3 sin x - 3 = 0 \Rightarrow 2 sin^{2} x - 3 sin x + 1 = 0$
2. Factorise in s = sin x:
  
  $(2 s - 1) (s - 1) = 0, s = \frac{1}{2}, s = 1$
3. sin x = 1/2 gives 30° and 150°; sin x = 1 gives only 90°, where α and 180° − α are the same angle:
  
  $x = 3 0^{\circ}, 15 0^{\circ}; x = 9 0^{\circ}$
Answer

Three solutions: x = 30°, 90° and 150°.

$x = 3 0^{\circ}, 9 0^{\circ}, 15 0^{\circ}$
Dividing to reach tan x 2 steps
Solve √3 sin x = cos x for −π ≤ x ≤ π.

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Show solution
1. Where cos x = 0, sin x = ±1 and the left side is ±√3, not 0, so no solution has cos x = 0 and dividing by it loses nothing:
  
  $\frac{3 sin x}{cos x} = \frac{cos x}{cos x} \Rightarrow tan x = \frac{1}{3}$
2. α = arctan(1/√3) = π/6, and tangent repeats every π:
  
  $x = \frac{π}{6}, x = \frac{π}{6} - π = - \frac{5 π}{6}$
Answer

x = −5π/6 or x = π/6.

$x = - \frac{5 π}{6}, x = \frac{π}{6}$
The general solution 3 steps
Find the general solution of cos 2x = −1/2 in radians, and list the solutions between 0 and 2π.

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Show solution
1. arccos(−1/2) = 2π/3, and cosine’s solutions come in pairs ±α, repeating every 2π:
  
  $2 x = \pm \frac{2 π}{3} + 2 nπ$
2. Divide by 2, which halves the period as well:
  
  $x = \pm \frac{π}{3} + nπ, n \in Z$
3. Between 0 and 2π, n = 0, 1 and 2 give
  
  $x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}$
Answer

x = ±π/3 + nπ for every integer n; between 0 and 2π these are π/3, 2π/3, 4π/3 and 5π/3.

$x = \pm \frac{π}{3} + nπ, n \in Z$

Supplementary problems, answers only

Solve cos x = 0.25 for 0° ≤ x ≤ 360°, to two decimal places.

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x = 75.52° or x = 284.48°. The calculator gives 75.52°, and the other value is −75.52°, which is 284.48° in the interval.

$x = 75.5 2^{\circ}, x = 284.4 8^{\circ}$
Solve tan(x/2) = 3 for 0° ≤ x ≤ 360°. How many solutions are there?

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One, x = 143.13°. The argument x/2 runs only from 0° to 180°, half a turn, and in that half turn tangent equals 3 only at x/2 = 71.57°.

$x = 143.1 3^{\circ}$
Solve 3sin²x − sin x − 2 = 0 for 0° ≤ x ≤ 360°, to two decimal places.

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(3sin x + 2)(sin x − 1) = 0. sin x = 1 gives 90°; sin x = −2/3 gives 221.81° and 318.19°. Three solutions.

$x = 9 0^{\circ}, 221.8 1^{\circ}, 318.1 9^{\circ}$
Solve 2sin²x = cos x + 1 for 0 ≤ x ≤ 2π, exactly.

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With sin²x = 1 − cos²x it becomes 2cos²x + cos x − 1 = 0, so cos x = 1/2 or cos x = −1: x = π/3, π and 5π/3.

$x = \frac{π}{3}, π, \frac{5 π}{3}$
Find the general solution of sin 3x = 1/2, in degrees.

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3x = 30° + 360°n or 3x = 150° + 360°n, so x = 10° + 120°n or x = 50° + 120°n.

$x = 1 0^{\circ} + 12 0^{\circ} n, x = 5 0^{\circ} + 12 0^{\circ} n$

Three rules, three symmetries of the circle 🖖

Sine is the height of the point on the circle, and the two points at height k are mirror images in the y-axis, which is why sin x = k gives α and 180° − α. Cosine is the horizontal coordinate, so its two points are mirror images in the x-axis: α and −α. Tangent is the slope of the line through O and the point, and that line meets the circle at two opposite points, α and α + 180°, so tangent repeats every half turn where sine and cosine need a whole one. Press cos x = −0.6 and the vertical line x = −0.6 cuts the circle at ±126.87°; press tan x = −2 and the line of slope −2 cuts it at 116.57° and 296.57°, half a turn apart.

The count comes before the answers 🖖

Press sin(2x − 30°) = √3/2. As x runs from 0° to 360°, the argument 2x − 30° runs from −30° to 690°, exactly two turns, and the spiral on the circle crosses each of the two solution rays once per turn: four solutions, counted before any of them is found. Solving sin u = √3/2 only between 0° and 360° finds u = 60° and 120°, so x = 45° and 75°, and stops two answers short. Over whole turns, counting one end and not the other, the count is exactly two per turn, for sine and cosine with −1 < k < 1 and for tangent with any k, since tangent’s period is half a turn. At k = ±1 the two points merge into one, and a solution that falls exactly on an end of the interval counts only if that end is ≤ rather than <.

Dividing by sin x deletes solutions 🖖

2 sin x cos x = sin x looks as if it simplifies to 2 cos x = 1, which gives 60° and 300° between 0° and 360°. But dividing by sin x assumes sin x ≠ 0, and at 0°, 180° and 360° both sides are 0, so those three angles solve the equation and the division threw them away. Factorising keeps them: sin x (2 cos x − 1) = 0, so sin x = 0 or cos x = 1/2, five solutions instead of two. Dividing is safe only when the divisor’s zeros cannot be solutions. In √3 sin x = cos x they cannot: where cos x = 0, sin x = ±1 and the left side is ±√3, so dividing by cos x to reach tan x = 1/√3 loses nothing.

Learning path

Trigonometric identities and equations

References (1)

Solving trigonometric equations on an interval and in general (adding 2πk to an initial solution), with multiple angles such as sin 2x, equations in quadratic form, and fundamental identities used first: J. Abramson, Precalculus 2e, OpenStax (2021), section 7.5, “Solving Trigonometric Equations”.

By the 42Math team · Reviewed against published sources. Corrections welcome.