Trigonometric Equation Solver

Solve sin, cos or tan equations over any interval and get every solution in range plus the general solution. Free.

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Summary

A trigonometric equation is solved by isolating the trig function, applying the inverse to get one angle, and then adding every other angle in the required interval.

Rearrange until a single trig function stands alone — factoring or substituting an identity if more than one appears — then take the inverse and use the function’s symmetry to find the remaining solutions.

The inverse function returns exactly one angle, but sine and cosine repeat every 360°360° and tangent every 180°180°, so a solution set without those periodic copies is incomplete rather than wrong.

01Definition

What is a trigonometric equation?

A trigonometric equation asks which angles make a statement true — sin⁡x=0.5\sin x = 0.5, or 2cos⁡2x−cos⁡x=02\cos^2 x - \cos x = 0. Unlike an identity, it is not true for every angle, so there is something genuine to solve.

The complication is that trig functions repeat. Sine returns to the same value every 360°360°, so if x=30°x = 30° works, so do 390°390°, 750°750°, and −330°-330° — infinitely many solutions. That is why questions almost always restrict the answer to an interval such as 0°≤x<360°0° \le x < 360° or [0,2π)[0, 2\pi).

The reliable method is always the same. Isolate the trig function, take the inverse to get the reference angle, decide which quadrants have the right sign, build the angles in those quadrants, and finally add or subtract full periods to sweep the whole interval.

Where you'll actually use this

  • Tide tables: finding the times of day when the water reaches a given height means solving a sinusoidal equation.
  • AC circuits: determining when the voltage in V=V0sin⁡(ωt)V = V_0\sin(\omega t) crosses a threshold.
  • Daylight modelling — which dates of the year have exactly 12 hours of sun.
  • Simple harmonic motion: when a pendulum or spring passes a given displacement.
  • Projectile launch angles, where a required range fixes sin⁡2θ\sin 2\theta and two launch angles both work.
Reference angle
The acute angle to the x-axis from which every quadrant solution is built.
Period
The interval after which a function repeats — 360°360° for sine and cosine, 180°180° for tangent.
ASTC
All-Sine-Tangent-Cosine — which functions are positive in quadrants I to IV.
General solution
Every solution expressed with an integer parameter, e.g. x=30°+360°nx = 30° + 360°n.
Principal value
The single angle an inverse trig function returns, within its restricted range.
02Formulas

The formulas you need

Keep these on hand — every method below builds on them.

Reference angle
α=∣inverse of the ratio’s absolute value∣\alpha = \left|\text{inverse of the ratio’s absolute value}\right|

The acute angle to the x-axis — every solution is built from it.

Sine positive
x=α  or  180°−αx = \alpha \;\text{or}\; 180° - \alpha

Quadrants I and II. Negative sine gives $180° + \alpha$ and $360° - \alpha$.

Cosine positive
x=α  or  360°−αx = \alpha \;\text{or}\; 360° - \alpha

Quadrants I and IV. Negative cosine gives $180° - \alpha$ and $180° + \alpha$.

Tangent positive
x=α  or  180°+αx = \alpha \;\text{or}\; 180° + \alpha

Quadrants I and III — tangent repeats every $180°$, not $360°$.

ASTC
All, Sine, Tangent, Cosine\text{All, Sine, Tangent, Cosine}

Which functions are positive in quadrants I–IV, counterclockwise.

General solution
x=α+360°n,n∈Zx = \alpha + 360°n, \quad n \in \mathbb{Z}

Used when no interval is given. For tangent the period is $180°$.

03Methods

How to solve trigonometric equations

Pick the method that fits the problem in front of you.

When to use it: The core method — every other technique reduces to this one at the end.
  1. 1Isolate the trig function so it stands alone: from 2sin⁡x=12\sin x = 1, get sin⁡x=12\sin x = \frac{1}{2}.
  2. 2Take the inverse of the absolute value to get the reference angle: α=sin⁡−1(0.5)=30°\alpha = \sin^{-1}(0.5) = 30°.
  3. 3Decide the quadrants from the sign — sine is positive, so quadrants I and II.
  4. 4Build each solution: x=30°x = 30° and x=180°−30°=150°x = 180° - 30° = 150°. Keep only those inside the interval.
04Worked examples

Trigonometric Equations, solved step by step

From easy to hard — pick a problem to see its full solution.

2sin⁡x=1,0°≤x<360°2\sin x = 1, \quad 0° \le x < 360°
  1. 1Isolate: sin⁡x=12\sin x = \frac{1}{2}.
  2. 2Reference angle: α=30°\alpha = 30°. Sine is positive, so quadrants I and II.
  3. 3Solutions: x=30°x = 30° and x=150°x = 150°.
05Practice

Now you try

Work each one on paper first, then check your answer.

Problem 1
sin⁡x=22,0°≤x<360°\sin x = \frac{\sqrt{2}}{2}, \quad 0° \le x < 360°
Problem 2
3tan⁡x+3=0,0°≤x<360°3\tan x + 3 = 0, \quad 0° \le x < 360°
Problem 3
2cos⁡x+3=0,0≤x<2π2\cos x + \sqrt{3} = 0, \quad 0 \le x < 2\pi
Problem 4
cos⁡2x−1=0,0°≤x<360°\cos^2 x - 1 = 0, \quad 0° \le x < 360°
Problem 5
tan⁡2x=1,0°≤x<180°\tan 2x = 1, \quad 0° \le x < 180°
06Common mistakes

Where points get lost

Each of these shows up on real graded work — and each has a simple fix.

Giving only the calculator’s answer and missing the other solutions in the interval.Inverse trig functions have a restricted range by design and return exactly one angle, which looks like the complete answer.Treat the calculator value as the reference angle only. Then ask which quadrants carry the right sign and build every solution in the interval from there.
Dividing both sides by a trig function: turning 2cos⁡2x=cos⁡x2\cos^2 x = \cos x into 2cos⁡x=12\cos x = 1.Cancelling a common factor is standard algebra, and it does make the equation shorter.Dividing by cos⁡x\cos x silently assumes it is non-zero, deleting the x=90°,270°x = 90°, 270° solutions. Move everything to one side and factor instead.
Dividing a multiple-angle solution by the coefficient too early.Solving for xx feels like the goal, so the division happens as soon as the first value appears.Widen the interval, find every solution for the full argument 2x2x across it, and divide only at the very end. Dividing first loses half the answers.
Answering in degrees when the interval was given in radians.Calculator mode and question notation drift apart, and π4\frac{\pi}{4} versus 45°45° both look correct in isolation.Read the interval first: [0,2π)[0, 2\pi) means radians throughout, 0°≤x<360°0° \le x < 360° means degrees. Set the calculator to match before the first keystroke.
Reporting a solution that lies outside the stated interval.Every quadrant is worked through mechanically, and the interval condition gets forgotten at the listing stage.Filter at the end: compare each candidate against the interval and discard the rest. Note that 0°≤x<360°0° \le x < 360° excludes 360°360° itself.

Photo Math Solver shows every intermediate step, so slips like these are easy to catch before they cost you marks.

07FAQ

Frequently asked questions

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