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Solving Trigonometric Equations: How to Find Every Solution in an Interval

A linear equation has one solution. A quadratic has two. A trigonometric equation has infinitely many, because sine and cosine repeat forever - and that single fact is behind most of the marks lost on this topic. Students find one solution, or two, and stop, when the question asked for every solution in an interval.

The good news is that the structure is completely regular. Each function has a fixed pattern of solutions inside one period, and everything else is that pattern shifted by whole periods. Once you can see it that way the work becomes mechanical.

This guide covers the general-solution formulas, the four situations that need extra care - quadratics, multiple angles, mixed functions, and squaring - and works five equations in full.

Why there are infinitely many solutions

Sine and cosine both have period 2π (360°): shifting the angle by a full turn returns you to the same point on the unit circle, so the coordinates are unchanged. Tangent has period π (180°), because the point diametrically opposite has both coordinates negated and the quotient y/x is therefore unaffected.

So if you find one angle satisfying sin θ = 1/2, you have really found infinitely many, spaced 2π apart in both directions. And there is a second family too: sine takes the same value at θ and at π − θ, because those two points on the unit circle have the same height.

That means every solvable sine equation has two families of solutions, not one. The same is true for cosine, where the mirror partner of θ is −θ. Tangent is the exception: it has a single family, spaced π apart, because its period already accounts for the symmetry.

sin⁡θ=k  ⟹  θ=arcsin⁡k+2πn   or   θ=π−arcsin⁡k+2πncos⁡θ=k  ⟹  θ=±arccos⁡k+2πntan⁡θ=k  ⟹  θ=arctan⁡k+πn\begin{aligned} \sin\theta = k &\;\Longrightarrow\; \theta = \arcsin k + 2\pi n \;\text{ or }\; \theta = \pi - \arcsin k + 2\pi n \\[4pt] \cos\theta = k &\;\Longrightarrow\; \theta = \pm\arccos k + 2\pi n \\[4pt] \tan\theta = k &\;\Longrightarrow\; \theta = \arctan k + \pi n \end{aligned}

n is any integer. In degrees, replace π with 180° and 2π with 360°.

  • sin θ = k and cos θ = k have no solutions at all when |k| > 1, since both functions are bounded between −1 and 1.
  • tan θ = k always has solutions, for every real k, because tangent takes every real value in each period.
  • Your calculator returns only the principal value - arcsin in [−90°, 90°], arccos in [0°, 180°], arctan in (−90°, 90°). Everything else you must supply yourself.

The procedure

Every problem of this type reduces to the same sequence. The discipline is in doing all five steps rather than stopping when you have an answer that looks plausible.

Step 5 is where the interval finally gets used, and it is worth doing systematically: take each family in turn, substitute n = 0, then n = 1, n = 2, then n = −1, and keep everything that lands inside the interval. Stop when the values leave it in both directions.

  • 1. Isolate a single trig function. Treat sin θ or cos θ as though it were the variable x and use ordinary algebra - factorising, collecting terms, taking a square root.
  • 2. If more than one function appears, use an identity to reduce it to one. The Pythagorean identity is the usual tool.
  • 3. Find the reference solution with the appropriate inverse function.
  • 4. Use the symmetry of the unit circle to produce every solution within one period, then attach the period to build the general solution.
  • 5. Restrict to the requested interval by substituting integer values of n.

Worked equations

The five below are ordered by the type of difficulty they introduce, not by length. Read the third and fifth carefully even if the first two look easy - they contain the mistakes that actually cost marks.

Equation 1 - the basic shape

Solve 2 sin θ − 1 = 0 for 0° ≤ θ < 360°.

  1. Isolate the sine.
    2sin⁡θ−1=0⟹sin⁡θ=122\sin\theta - 1 = 0 \quad\Longrightarrow\quad \sin\theta = \tfrac{1}{2}
  2. Find the reference angle from the first quadrant.
    θref=arcsin⁡12=30∘\theta_{\text{ref}} = \arcsin\tfrac{1}{2} = 30^{\circ}
  3. Sine is positive in Quadrants I and II, so there are two solutions in one turn: the reference angle itself, and its supplement.
    θ=30∘andθ=180∘−30∘=150∘\theta = 30^{\circ} \qquad\text{and}\qquad \theta = 180^{\circ} - 30^{\circ} = 150^{\circ}

    Returning only 30° is the single most common error on this topic. The calculator gives one value; the circle gives two.

  4. The general solution adds full turns to each family.
    θ=30∘+360∘norθ=150∘+360∘n\theta = 30^{\circ} + 360^{\circ}n \quad\text{or}\quad \theta = 150^{\circ} + 360^{\circ}n
  5. Restrict to [0°, 360°): only n = 0 lands inside the interval for either family.

Answer. θ = 30° and θ = 150°.

Equation 2 - quadratic in cosine

Solve 2 cos²θ + cos θ − 1 = 0 for 0 ≤ θ < 2π.

  1. Only cosine appears, so substitute u = cos θ and treat it as an ordinary quadratic.
    2u2+u−1=0⟹(2u−1)(u+1)=02u^{2} + u - 1 = 0 \quad\Longrightarrow\quad (2u - 1)(u + 1) = 0
  2. Read off the two values of u, and remember that both are constraints on cos θ.
    cos⁡θ=12orcos⁡θ=−1\cos\theta = \tfrac{1}{2} \qquad\text{or}\qquad \cos\theta = -1
  3. Solve the first. Cosine is positive in Quadrants I and IV, and arccos(1/2) = π/3.
    θ=π3orθ=2π−π3=5π3\theta = \dfrac{\pi}{3} \qquad\text{or}\qquad \theta = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}
  4. Solve the second. cos θ = −1 happens at the single point (−1, 0).
    θ=π\theta = \pi

    Values of ±1 are the boundary cases where the two families collapse into one solution per period. Do not double-count them.

  5. Collect everything in [0, 2π).

Answer. θ = π/3, π, and 5π/3.

Equation 3 - a multiple angle, and the interval trap

Solve sin 2θ = √3/2 for 0° ≤ θ < 360°.

  1. Substitute u = 2θ. This is the step that makes the problem safe, because the interval must be transformed as well.
    0∘≤θ<360∘⟹0∘≤u<720∘0^{\circ} \le \theta < 360^{\circ} \quad\Longrightarrow\quad 0^{\circ} \le u < 720^{\circ}

    Solving for θ inside [0°, 360°) directly is how people lose half the answers. Doubling the angle doubles the number of solutions.

  2. Solve for u across the doubled interval. The reference angle is arcsin(√3/2) = 60°, and sine is positive in Quadrants I and II.
    u=60∘, 120∘in the first turnu = 60^{\circ},\ 120^{\circ} \quad\text{in the first turn}
  3. Add 360° to each to pick up the second turn, which is still inside [0°, 720°).
    u=60∘, 120∘, 420∘, 480∘u = 60^{\circ},\ 120^{\circ},\ 420^{\circ},\ 480^{\circ}
  4. Halve every value to return to θ.
    θ=30∘, 60∘, 210∘, 240∘\theta = 30^{\circ},\ 60^{\circ},\ 210^{\circ},\ 240^{\circ}
  5. Check one: sin(2 × 210°) = sin 420° = sin 60° = √3/2. Correct.

Answer. θ = 30°, 60°, 210°, 240° - four solutions, not two.

Equation 4 - two different functions, resolved with an identity

Solve 2 sin²θ + 3 cos θ − 3 = 0 for 0 ≤ θ < 2π.

  1. Sine and cosine both appear, so nothing can be isolated yet. Use the Pythagorean identity to eliminate the squared term - always convert the squared function, never the linear one.
    sin⁡2θ=1−cos⁡2θ\sin^{2}\theta = 1 - \cos^{2}\theta
  2. Substitute and expand.
    2(1−cos⁡2θ)+3cos⁡θ−3=0⟹−2cos⁡2θ+3cos⁡θ−1=02(1 - \cos^{2}\theta) + 3\cos\theta - 3 = 0 \quad\Longrightarrow\quad -2\cos^{2}\theta + 3\cos\theta - 1 = 0
  3. Multiply through by −1 and factorise.
    2cos⁡2θ−3cos⁡θ+1=0⟹(2cos⁡θ−1)(cos⁡θ−1)=02\cos^{2}\theta - 3\cos\theta + 1 = 0 \quad\Longrightarrow\quad (2\cos\theta - 1)(\cos\theta - 1) = 0
  4. Solve each factor.
    cos⁡θ=12  ⟹  θ=π3, 5π3cos⁡θ=1  ⟹  θ=0\cos\theta = \tfrac{1}{2} \;\Longrightarrow\; \theta = \dfrac{\pi}{3},\ \dfrac{5\pi}{3} \qquad \cos\theta = 1 \;\Longrightarrow\; \theta = 0
  5. Verify θ = 0 in the original equation: 2(0)² + 3(1) − 3 = 0. Correct.

Answer. θ = 0, π/3, and 5π/3.

Equation 5 - squaring, and the extraneous solutions it creates

Solve sin θ − cos θ = 1 for 0 ≤ θ < 2π.

  1. No factorisation or identity reduces this directly, so square both sides - a legal move on an equation, unlike on an identity, but one that can add solutions.
    (sin⁡θ−cos⁡θ)2=12⟹sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ=1(\sin\theta - \cos\theta)^{2} = 1^{2} \quad\Longrightarrow\quad \sin^{2}\theta - 2\sin\theta\cos\theta + \cos^{2}\theta = 1
  2. The two squares combine to 1 by the Pythagorean identity, and the middle term is the sine double-angle formula.
    1−sin⁡2θ=1⟹sin⁡2θ=01 - \sin 2\theta = 1 \quad\Longrightarrow\quad \sin 2\theta = 0
  3. Solve for 2θ over the doubled interval [0, 4π).
    2θ=0, π, 2π, 3π⟹θ=0, π2, π, 3π22\theta = 0,\ \pi,\ 2\pi,\ 3\pi \quad\Longrightarrow\quad \theta = 0,\ \dfrac{\pi}{2},\ \pi,\ \dfrac{3\pi}{2}
  4. Now test every candidate in the original equation, because squaring does not preserve sign.
    θ=0:0−1=−1≠1  rejectθ=π2:1−0=1  ✓θ=π:0−(−1)=1  ✓θ=3π2:−1−0=−1≠1  reject\begin{aligned} \theta = 0: &\quad 0 - 1 = -1 \neq 1 \;\text{reject} \\ \theta = \tfrac{\pi}{2}: &\quad 1 - 0 = 1 \;\checkmark \\ \theta = \pi: &\quad 0 - (-1) = 1 \;\checkmark \\ \theta = \tfrac{3\pi}{2}: &\quad -1 - 0 = -1 \neq 1 \;\text{reject} \end{aligned}

    Two of the four candidates satisfy sin θ − cos θ = −1 instead. Squaring made those indistinguishable, which is exactly why the check is compulsory.

Answer. θ = π/2 and θ = π. The other two candidates are extraneous.

Never divide by a trig function

One more trap deserves its own section, because it destroys solutions silently rather than adding ones you can catch.

Consider sin θ cos θ = sin θ. It is tempting to divide both sides by sin θ to get cos θ = 1. That division is only valid when sin θ ≠ 0, and by performing it you have thrown away every solution where sin θ = 0 without noticing.

The correct move is always to bring everything to one side and factorise. Here that gives sin θ (cos θ − 1) = 0, so either sin θ = 0, giving θ = 0 and θ = π on [0, 2π), or cos θ = 1, giving θ = 0 again. The full solution set is {0, π} - and the θ = π you would have lost is a genuine answer.

The same principle applies to cancelling a common factor anywhere in this topic. Factor and set each factor to zero; never divide it away.

Reading the interval correctly

Intervals are stated in several conventions and each changes the answer set, so read the notation before you start.

Square brackets include the endpoint and round brackets exclude it, so [0, 2π) contains 0 but not 2π. Because 0 and 2π are the same point on the circle, including both would list one solution twice - which is precisely why the half-open form is standard.

Watch for intervals that are not one full period. "All solutions in [0, π]" needs the same general solution, just filtered more aggressively. And when the equation involves a shifted argument such as cos(θ − π/3) or a compressed one such as sin 3θ, transform the interval along with the substitution, exactly as in Equation 3 - for sin 3θ on [0, 2π) you are solving for u = 3θ on [0, 6π), which typically yields six solutions.

  • Degrees or radians: match whatever the question uses, and set your calculator accordingly.
  • "General solution" means give the families with + 2πn or + πn, not a finite list.
  • "Principal solution" or "principal value" means the single value the inverse function returns.
  • If the equation contains kθ, the interval for kθ is k times as wide, and you should expect roughly k times as many solutions.

Frequently asked questions

How many solutions should I expect in [0, 2π)?

For sin θ = k or cos θ = k with |k| < 1, exactly two. For |k| = 1, exactly one. For tan θ = k, exactly two, because tangent completes two periods in a full turn. A quadratic that factors into two such equations can give up to four, and an equation in kθ multiplies the count by roughly k.

Why does my calculator only give one answer?

Inverse trig functions have to be single-valued, so their ranges are restricted: arcsin to [−90°, 90°], arccos to [0°, 180°], arctan to (−90°, 90°). The calculator returns the principal value from that range. Producing the second solution in the period, and then adding whole periods, is your job.

When do I have to check my answers?

Always after squaring both sides, always after using an identity that is undefined somewhere - anything involving tan, sec, cot or csc can introduce a candidate where the original expression does not exist - and always if you have multiplied by an expression containing the variable. Substituting back into the original equation takes seconds.

What if the equation has no exact-value solution, like sin θ = 0.3?

The method is identical; only the arithmetic changes. arcsin(0.3) ≈ 17.46°, so the two solutions in [0°, 360°) are approximately 17.46° and 180° − 17.46° = 162.54°. Keep the extra decimal places in intermediate steps and round only at the end.

Is there a way to check a whole solution set quickly?

Substitute each answer back into the original equation - that catches extraneous solutions. To catch missing ones, count against what you expected: two per period per branch for a simple sine or cosine equation, multiplied up if the argument is kθ. If you have fewer than expected, you probably divided by a trig function somewhere.

Keep going

sin²θ + cos²θ = 1sin 2θ = 2 sin θ cos θarcsin and its rangearccos and its rangearctan and its rangeSine is oddTopic: InverseTopic: Double AngleGuide: memorise the unit circleGuide: how to prove a trig identityAll guides

Practise it

Reading a method is not the same as running it under time pressure. The workspace drills this material in the quiz, and the triangle solver shows a worked solution for any triangle you type in.

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