Salam Academy
Sign inCreate account
Learn7357Paper 1 · PureTrigonometrySolving trigonometric equations
Trigonometry · Subtopic 12 of 12

Solving Trigonometric Equations

How to find every solution of a trig equation in an interval: principal value, symmetry, and period. Then equations with 2x2x or x+30∘x+30^\circ inside, quadratics in sin⁡x\sin x or cos⁡x\cos x, and equations that need an identity first.

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 55 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Find all solutions of sin⁡x=k\sin x=k, cos⁡x=k\cos x=k and tan⁡x=k\tan x=k in a given interval, in degrees or radians.
  • Solve equations such as sin⁡2x=k\sin2x=k and tan⁡(x−30∘)=k\tan(x-30^\circ)=k by adjusting the interval.
  • Solve quadratic equations in sin⁡x\sin x, cos⁡x\cos x or tan⁡x\tan x.
  • Use tan⁡x≡sin⁡xcos⁡x\tan x\equiv\dfrac{\sin x}{\cos x} and sin⁡2x+cos⁡2x≡1\sin^2x+\cos^2x\equiv1 to get an equation in one function.
Pause and think

Your calculator says sin⁡−10.5=30∘\sin^{-1}0.5=30^\circ. Using the graph of y=sin⁡xy=\sin x, where else between 0∘0^\circ and 360∘360^\circ does sin⁡x=0.5\sin x=0.5?

Section 2 of 14

Core idea

Start hereWhen is the tide exactly 3 m?

A harbour's water depth follows a sine wave. When is it exactly 33 m? Because the wave repeats, the answer isn't one time but several each day. Trig equations have many solutions; the skill is finding all of them in a given range, systematically.

The big idea

sin⁡x=k\sin x=k (with −1≤k≤1-1\le k\le1) has two solutions per cycle. Get one from the calculator, the second from symmetry (180∘−α180^\circ-\alpha for sin, 360∘−α360^\circ-\alpha for cos, α+180∘\alpha+180^\circ for tan), then add or subtract whole periods to fill the range.

1. The second solution

sin x = 0.5 has two solutions in one cycle
50100150200250300350−1−0.50.510x (degrees)yy = sin xy = 0.5symmetry30°150°
The sine curve is symmetric about 90∘90^\circ, so the second solution is 180∘−30∘=150∘180^\circ-30^\circ=150^\circ.
sin⁡x=k:x=α, 180∘−α(+360∘n)cos⁡x=k:x=α, 360∘−α(+360∘n)tan⁡x=k:x=α(+180∘n)\begin{array}{lll}\sin x=k:&x=\alpha,\ 180^\circ-\alpha&(+360^\circ n)\\\cos x=k:&x=\alpha,\ 360^\circ-\alpha&(+360^\circ n)\\\tan x=k:&x=\alpha&(+180^\circ n)\end{array}
where
α\alpha
the principal value from the calculator (sin⁡−1k\sin^{-1}k etc.).
nn
any whole number.
Solutions of sin x = k as k changes
50100150200250300350−1−0.50.510x (degrees)y
For negative kk the calculator gives a negative angle; add 360∘360^\circ to bring it into range and use symmetry about 270∘270^\circ for the other.
Worked examplesin⁡x=−0.4\sin x=-0.4, 0≤x≤360∘0\le x\le360^\circ

Solve, to 1 d.p.

Goal
All solutions in range.
Strategy
Principal value, symmetry, then shift into range.
  1. α=sin⁡−1(−0.4)=−23.6∘\alpha=\sin^{-1}(-0.4)=-23.6^\circ
    Outside the range, but still useful.
  2. 180∘−(−23.6∘)=203.6∘180^\circ-(-23.6^\circ)=203.6^\circ
    Symmetry of sine.
  3. −23.6∘+360∘=336.4∘-23.6^\circ+360^\circ=336.4^\circ
    Add a period.
Answerx=203.6∘x=203.6^\circ or 336.4∘336.4^\circ.
✓ CheckSine is negative in Q3 and Q4 ✓; sin⁡203.6∘=−0.400\sin203.6^\circ=-0.400 ✓.

2. Multiple and shifted angles

Worked examplecos⁡2x=0.5\cos2x=0.5, 0≤x≤360∘0\le x\le360^\circ

Solve.

Goal
All xx.
Strategy
Let u=2xu=2x; the range for uu doubles; solve for uu then halve.
  1. 0≤u≤720∘0\le u\le720^\circ
    Multiply the range by 22.
  2. u=60∘,300∘,420∘,660∘u=60^\circ,300^\circ,420^\circ,660^\circ
    α=60∘\alpha=60^\circ, 360−60360-60, then +360∘+360^\circ each.
  3. x=30∘,150∘,210∘,330∘x=30^\circ,150^\circ,210^\circ,330^\circ
    Halve.
AnswerFour solutions.
✓ Checkcos⁡(2×150∘)=cos⁡300∘=0.5\cos(2\times150^\circ)=\cos300^\circ=0.5 ✓.
Worked exampletan⁡(x−30∘)=1\tan(x-30^\circ)=1, 0≤x≤360∘0\le x\le360^\circ

Solve.

Goal
All xx.
Strategy
u=x−30∘u=x-30^\circ; shift the range.
  1. −30∘≤u≤330∘-30^\circ\le u\le330^\circ
    Subtract 30∘30^\circ from both ends.
  2. u=45∘,225∘u=45^\circ,225^\circ
    Tan period 180∘180^\circ.
  3. x=75∘,255∘x=75^\circ,255^\circ
    Add 30∘30^\circ.
Answer75∘75^\circ, 255∘255^\circ.
✓ Checktan⁡45∘=1\tan45^\circ=1 ✓.
Common misconceptionForgetting to change the range For cos⁡2x\cos2x, solving only 0≤2x≤360∘0\le2x\le360^\circ loses half the answers.

3. Quadratic trig equations

Worked examplesin⁡xcos⁡x=sin⁡x\sin x\cos x=\sin x, 0≤x≤360∘0\le x\le360^\circ

Solve.

Goal
All xx.
Strategy
Rearrange to zero and factorise. Never divide by sin⁡x\sin x (it may be zero).
  1. sin⁡xcos⁡x−sin⁡x=0\sin x\cos x-\sin x=0
    Subtract.
  2. sin⁡x(cos⁡x−1)=0\sin x(\cos x-1)=0
    Factorise.
  3. sin⁡x=0⇒x=0∘,180∘,360∘\sin x=0\Rightarrow x=0^\circ,180^\circ,360^\circ
    First factor.
  4. cos⁡x=1⇒x=0∘,360∘\cos x=1\Rightarrow x=0^\circ,360^\circ
    Already listed.
Answerx=0∘,180∘,360∘x=0^\circ,180^\circ,360^\circ.
✓ CheckDividing by sin⁡x\sin x would have given only cos⁡x=1\cos x=1 and lost 180∘180^\circ.
Solving a trig equation

Factorise rather than divide by a trig function.

Pause and think

Solve 3tan⁡2x=13\tan^2x=1, 0≤x≤2π0\le x\le2\pi.

Summary

  • Principal value, symmetry, periods.
  • Adjust ranges for kxkx and x±cx\pm c.
  • Factorise; never cancel.
Key words
Principal value
The calculator answer.
General solution
All solutions, written with +360n or +180n.
HistoryTables before calculators. Until the 1970s, students found sin⁡−10.6\sin^{-1}0.6 by looking it up in printed tables. The first great sine table was compiled around 150 BC by Hipparchus, and al-Battani and Ulugh Beg later produced tables accurate to many decimal places, used by astronomers for centuries.
Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free accountSign in