Why does a trig equation have more than one solution?

Because trig functions are periodic — sine and cosine repeat every 360°, tangent every 180°. Any angle that works has infinitely many partners, which is why questions restrict the answer to a stated interval.

What is a reference angle?

The acute angle between the terminal side and the x-axis. It is what the inverse function gives you from the absolute value of the ratio, and every solution in every quadrant is constructed from it.

How do I know which quadrants to use?

From the sign of the ratio, using ASTC: All positive in quadrant I, Sine in II, Tangent in III, Cosine in IV. If \sin x is negative, you want the two quadrants where sine is not positive — III and IV.

How do I handle \sin 2x or \cos 3x?

Multiply the interval by the same coefficient, solve for the whole argument across that wider range, then divide each solution by the coefficient at the end. A 2x equation has twice as many solutions as the same equation in x.

Can I divide both sides by \sin x?

No — that throws away every solution where \sin x = 0. Move all terms to one side and factor instead, which keeps both branches and costs nothing.

What if the question does not give an interval?

Then it wants the general solution, written with an integer parameter: x = 30° + 360°n or x = 150° + 360°n for n \in \mathbb{Z}. For tangent the period is 180°, so a single family covers it.

You can start solving trigonometric equations for free by adding the Chrome extension — try it on your own worksheet before deciding if you need anything more.

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Photo Math Solver Solved

2\sin x = 1, \quad 0° \le x < 360°

1Isolate: $\sin x = \frac{1}{2}$ .
2Reference angle: $\alpha = 30°$ . Sine is positive, so quadrants I and II.
3Solutions: $x = 30°$ and $x = 150°$ .

On this page01Definition 02Formulas 03Methods 04Examples 05Practice 06Mistakes 07FAQ

01Definition

What is a trigonometric equation?

A trigonometric equation asks which angles make a statement true — $\sin x = 0.5$ , or $2\cos^2 x - \cos x = 0$ . 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 $360°$ , so if $x = 30°$ works, so do $390°$ , $750°$ , and $-330°$ — infinitely many solutions. That is why questions almost always restrict the answer to an interval such as $0° \le x < 360°$ or $[0, 2\pi)$ .

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 $V = V_0\sin(\omega t)$ 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 $\sin 2\theta$ 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 — $360°$ for sine and cosine, $180°$ 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. $x = 30° + 360°n$ .
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

\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 = \alpha \;\text{or}\; 180° - \alpha

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

Cosine positive

x = \alpha \;\text{or}\; 360° - \alpha

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

Tangent positive

x = \alpha \;\text{or}\; 180° + \alpha

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

ASTC

\text{All, Sine, Tangent, Cosine}

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

General solution

x = \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.

1Isolate the trig function so it stands alone: from $2\sin x = 1$ , get $\sin x = \frac{1}{2}$ .
2Take the inverse of the absolute value to get the reference angle: $\alpha = \sin^{-1}(0.5) = 30°$ .
3Decide the quadrants from the sign — sine is positive, so quadrants I and II.
4Build each solution: $x = 30°$ and $x = 180° - 30° = 150°$ . 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.

2\sin x = 1, \quad 0° \le x < 360°

1Isolate: $\sin x = \frac{1}{2}$ .
2Reference angle: $\alpha = 30°$ . Sine is positive, so quadrants I and II.
3Solutions: $x = 30°$ and $x = 150°$ .

Try it on your own trigonometric equations homework

Photo Math Solver reads a trigonometric equation straight from a screenshot, interval and all, and shows the full method: isolate the trig function, find the reference angle, work out which quadrants match the sign, and list every solution in range. It is the "and also" solutions that cost marks, and those are exactly the ones an inverse-function keystroke on a calculator will not give you.

05Practice

Now you try

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

Problem 1

\sin x = \frac{\sqrt{2}}{2}, \quad 0° \le x < 360°

Problem 2

3\tan x + 3 = 0, \quad 0° \le x < 360°

Problem 3

2\cos x + \sqrt{3} = 0, \quad 0 \le x < 2\pi

Problem 4

\cos^2 x - 1 = 0, \quad 0° \le x < 360°

Problem 5

\tan 2x = 1, \quad 0° \le x < 180°

Stuck on your trigonometric equations homework? Screenshot it

The interval is as much a part of a trig equation as the equation itself, and it is usually set in small type beside it. Screenshot both together with Photo Math Solver and the solution works within the range you were actually given — every solution in it, none outside it, in the right units.

06Common mistakes

Where points get lost

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

The mistake	Why it happens	The 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 $2\cos^2 x = \cos x$ into $2\cos x = 1$ .	Cancelling a common factor is standard algebra, and it does make the equation shorter.	Dividing by $\cos x$ silently assumes it is non-zero, deleting the $x = 90°, 270°$ solutions. Move everything to one side and factor instead.
Dividing a multiple-angle solution by the coefficient too early.	Solving for $x$ 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 $2x$ 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 $\frac{\pi}{4}$ versus $45°$ both look correct in isolation.	Read the interval first: $[0, 2\pi)$ means radians throughout, $0° \le x < 360°$ 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 $0° \le x < 360°$ excludes $360°$ itself.

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

07FAQ