For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: compound and double-angle identities, auxiliary-angle forms, equation solving in an interval, identity proofs, and modelling form the main route.
School-sensitive extension: general solutions, proof-heavy inequalities, and multi-stage models may vary by school.
2027 national comparison: K341 Topic G1 includes compound and double-angle formulas, auxiliary-angle expressions, simplification, equations in a stated interval, identity proofs, and models.
Check your school: confirm interval notation, accepted auxiliary-angle conventions, and whether general solutions are assessed.
Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 13) Trigonometry II cover? A: Trig equations, compound angles, and auxiliary angle method for IP AMaths.
With radian fluency in hand, solve trig equations by mapping solutions across quadrants and applying compound-angle identities.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
The core idea is simple: Trigonometry II is about solving equations across the correct interval.
Use it as a working check: Reduce the equation to a familiar trig function, find all quadrant solutions, then check the requested range in radians.
Then go one layer deeper: Example: if sin x = 1/2 from 0 to just before 2pi, include both pi/6 and 5pi/6, not just the calculator's first answer.
Solving route for trig equations
The goal is not just to solve one angle. It is to find every angle in the stated interval.
Question cue
First move
Check before final answer
Quadratic in sinx, cosx, or tanx
Substitute a temporary variable, solve the algebra, then convert back to trig equations.
Reject algebra roots outside the possible trig range, such as sinx=2.
Equation contains sin2x, cos2x, or tan2x
Decide whether to expand the double angle or solve for 2x first.
If solving for 2x
Equation has asinx+bcosx
Rewrite as an auxiliary angle form before solving.
Account for both branches of the sine or cosine equation.
Equation has products of trig factors
Factor first, then solve each factor separately.
Do not divide by a trig factor that might be zero.
Common trap: a calculator principal value is a starting point, not the full answer. Use quadrant signs and the interval endpoints to list every valid solution.
Equation route checkpoint
Before solving, decide whether the equation should be factorised, rewritten with an identity, or converted to auxiliary-angle form.
Equation shape
Safer first move
Why it works
Trap to avoid
Product form, such as cosx(2sinx−1)=0
Set each factor to zero.
Each factor can create a separate family of angles.
Dividing by cosx and losing cosx=0 solutions.
Double-angle form, such as sin2x=cosx
Expand if a common factor appears.
Expansion can reveal a factorisation route.
Solving only for 2x and forgetting the original interval for x.
Linear combination, such as 3sinx+4cosx=2
Convert to Rsin(x+α) or Rcos(x−α)
Quadratic in one trig function
Substitute temporarily, then convert back.
The algebra step is separate from the angle step.
Accepting impossible roots such as sinx=2.
Worked check: from sin2x=cosx, expanding gives 2sinxcosx=cosx, so cosx(2sinx−1)=0. This gives the cosx=0 branch and the sinx=21 branch. If you divide by cosx, the two right-angle solutions disappear.
Misconception check: cancelling a trig factor is only safe if you have already handled the case where that factor is zero.
1 Identities to deploy
Compound: sin(A±B)=sinAcosB±cosAsinB.
Compound: cos(A±B)=cosAcosB∓sinAsinB.
Auxiliary angle: express asinx+bcosx=Rsin(x+α), where R=a2+b2
2 Worked example - Solve trig equation
Solve 2sin2x−3sinx+1=0 for 0≤x<2π.
Let y=sinx. Equation becomes 2y2−3y+1=0.
Factor: (2y−1)(y−1)=0.
So y=21 or y=1.
If sinx=21, then x=6π
If sinx=1, then x=2π.
Solutions: x=6π,2π,65π.
3 Worked example - Auxiliary angle
Solve 3sinx+4cosx=2 for 0≤x<2π.
Compute R=32+42=5.
Choose α so that cosα=53 and sinα=54
Rewrite: 3sinx+4cosx=5sin(x+α).
Equation becomes sin(x+α)=52.
Principal solution: x+α=arcsin(52).
General solutions within range: x+α=arcsin(52) or x+α=π−arcsin(52)
Subtract α to get x. Numerically, α=0.927 rad, arcsin(52)=0.411
Hence x≈−0.516 rad; If we add 2π → 5.767 rad. Or take the second branch x≈1.803 rad.
Auxiliary-angle interval checkpoint
When an equation is rewritten as Rsin(x+α)=k, solve for the whole angle x+α first. The allowed interval for x+α is the original interval for x, shifted by α.
Step
What to write
Common trap
Shift the interval
If 0≤x<2π, then α≤x+α<2π+α.
Solving only in 0≤x+α<2π and missing an endpoint solution.
Solve the sine or cosine equation
List every value of x+α inside the shifted interval.
Keeping only the calculator's principal angle.
Subtract α
Convert each whole-angle solution back to x.
Forgetting to subtract α from every branch.
Check the original interval
Keep only answers that satisfy the original range for x.
Leaving a negative value such as −0.516 without adding 2π when the range is 0≤x<2π
Worked check: in the example above, α≈0.927, so x+α must satisfy 0.927≤x+α<7.210. The first sine branch gives x+α≈0.411, which is below this shifted interval. Add 2π to get 6.694, then subtract α to obtain x≈5.767. The second branch gives x+α≈2.731, so x≈1.803.
Misconception check: adding 2π is not a way to change the answer after the fact. It represents the same sine value in the correct shifted interval before converting back to x.
4 Worked example - Using double-angle identities in equations
Solve sin(2x)=cos(x) for 0≤x<2π.
Apply the double-angle identity sin(2x)=2sin(x)cos(x).
Equation becomes 2sin(x)cos(x)=cos(x).
Factor to cosx(2sinx−1)=0.
Either cosx=0, giving x=2π or x=23π
For sinx=21, we have x=6π
Answer: x=6π,2π,65π,23π.
5 Practice Quiz
Stress-test compound-angle identities, auxiliary-angle rewriting, and equation solving so you can move quickly in timed papers.
6 Try this
Solve 2cos2x=3 for 0≤x<2π, displaying answers in exact form.
, double the interval before dividing answers by
2
.
.
One trig function is easier to solve across the interval.
