standarddeviationcalculator.net

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Trigonometric equation solver

Type an equation in x (or θ) and get every solution on the interval, exact where possible. Equations that reduce to sin, cos or tan of one angle get the substitution worked out and the general solution; anything else is solved numerically and graphed.

Examples: sin x = 0.5, tan(2x) = 1, sin x + cos x = 1, 2sin^2(x) = 1 − cos x. Use x or θ.

Solutions in [0, 2π) x = 0, 2π/3, 4π/3
Number of solutions3
Interval[0, 2π)
General solution (n any integer)x = 2πn
x = 2π/3 + 2πn
x = −2π/3 + 2πn
Period of the equation2π
In degrees0°, 120°, 240°
Check: both sides agree at each solution
xLeft sideRight side
000
2π/300
4π/300
0π/2π3π/22π-1-0.500.511.52 x

━ y = 2(cos(x))^2 - cos(x) - 1   ┄ y = 0

Each solution is where the two curves meet.

Show the working, step by step

  1. Read the equation (as the calculator parsed it).

    2(cos(x))^2 - cos(x) - 1 = 0

  2. Only one trig expression appears, cos x. Let u = cos(x) and rewrite.

    2u² − u − 1 = 0

  3. Solve for u.

    u = 1 or u = −1/2

  4. Solve each basic equation for the angle.

    cos x = 1: principal value 0, so x = 2πn cos x = −1/2: principal value 2π/3, so x = 2π/3 + 2πn or x = −2π/3 + 2πn

  5. Pick the solutions that fall in [0, 2π).

    0, 2π/3, 4π/3

How to type equations

Write the functions the way you would on paper: sin x, cos(2x), tan²x or tan^2(x), sec x, arcsin(x). Use pi or π, √3 or sqrt(3). sin x/2 is read as (sin x)/2; write sin(x/2) for the sine of half x. Leave the interval blank for one full turn starting at 0, or type both ends (for example −pi and pi) for a closed interval.

The basic equations

Everything reduces to one of three forms, with α the principal value from the inverse function:

sin x = k: x = α + 2πn or x = π − α + 2πn, α = arcsin k cos x = k: x = ±α + 2πn, α = arccos k tan x = k: x = α + πn, α = arctan k

sin x = k and cos x = k need −1 ≤ k ≤ 1; tan x = k works for any k. In degrees replace π by 180° and 2π by 360°.

A worked example

The default equation is 2cos²x − cos x − 1 = 0 on [0, 2π).

  1. Only cos x appears, so let u = cos x: 2u² − u − 1 = 0.
  2. Factor: (2u + 1)(u − 1) = 0, so u = 1 or u = −1/2.
  3. cos x = 1 gives x = 2πn, which is 0 in the interval.
  4. cos x = −1/2: arccos(−1/2) = 2π/3, so x = ±2π/3 + 2πn, which is 2π/3 and 4π/3 in the interval.
  5. Solutions: x = 0, 2π/3, 4π/3 (0°, 120°, 240°).

The graph under the result plots each side as a curve; the solutions are where they meet, and the table substitutes each one back into both sides as a check.

More examples

EquationSolutions in [0, 2π)General solution
sin x = 0.5π/6, 5π/6π/6 + 2πn, 5π/6 + 2πn
tan 2x = 1π/8, 5π/8, 9π/8, 13π/8π/8 + n·π/2
sin x + cos x = 10, π/22πn, π/2 + 2πn
sin 3x = −√3/24π/9, 5π/9, 10π/9, 11π/9, 16π/9, 17π/9−π/9 + n·2π/3, 4π/9 + n·2π/3
sin x = 2nonenone (sine never exceeds 1)

sin x + cos x = 1 mixes two functions. By hand you would write the left side as √2·sin(x + π/4) and solve sin(x + π/4) = 1/√2. The solver instead finds the crossings numerically, confirms that the equation repeats every 2π, and builds the general solution from the solutions in one period.

Common mistakes

  • Stopping at the calculator’s one answer. arcsin, arccos and arctan return one angle; the second family comes from the symmetry of the unit circle.
  • Dividing by a trig function. Factor instead, or you lose the solutions where it is zero.
  • Forgetting the multiple angle. For sin 2x = k, solve for 2x over twice the interval, then halve.
  • Squaring both sides. This can add false solutions; check each answer in the original equation, as the table here does.

Common questions

How do I solve sin x = 0.5?

arcsin 0.5 = π/6 (30°). Sine is also positive in the second quadrant, so π − π/6 = 5π/6 (150°) is the other solution in one turn. The general solution is x = π/6 + 2πn or x = 5π/6 + 2πn, for any integer n.

What is a general solution?

A formula for every solution, not just those in one interval. Trig functions repeat, so once you have the solutions in one period you add whole periods: 2π (360°) for sin and cos, π (180°) for tan. For tan 2x = 1 the period is halved to π/2, giving x = π/8 + n·π/2.

Why does tan(2x) = 1 have four solutions in [0, 2π)?

As x runs from 0 to 2π, 2x runs from 0 to 4π, twice round the circle. tan θ = 1 has two solutions per turn, so there are four: x = π/8, 5π/8, 9π/8 and 13π/8. Whenever the angle is kx, expect k times as many solutions.

Can it solve equations with no algebraic method, like sin x = x/2?

Yes. Anything the solver cannot reduce to a basic equation is solved numerically: it scans the interval for places where the two sides cross and refines each one. On [−π, π], sin x = x/2 has three solutions: x = 0 and x = ±1.89549. Set the interval yourself for equations like this, since they have no repeating pattern.

Should I divide both sides by sin x?

No: you lose the solutions where sin x = 0. Factor instead. 2sin²x − sin x = 0 becomes sin x (2sin x − 1) = 0, so sin x = 0 or sin x = 1/2, giving 0°, 30°, 150° and 180° in [0°, 360°). Dividing by sin x would have kept only 30° and 150°.

Written and reviewed by our editorial team. Last updated . Method and sources: how these numbers are computed.