Solving trig equations in given intervals, finding principal and secondary solutions.
Answer at least 3 of 3 correctly to complete this section.
Trigonometric functions are periodic — they repeat. So unlike x2=4 (which has two solutions), an equation like sinθ=21 has infinitely many solutions. The skill at A-Level is finding all solutions in a given range, and the CAST diagram is your primary tool for this.
This topic brings together everything you know about trig: exact values, identities, and algebraic manipulation. It appears on virtually every A-Level paper, often as a 5–6 mark question that requires you to use an identity first before solving.
When you press sin−1 on your calculator, you get one answer — the principal value. The ranges are:
| Function | Principal value range |
|---|---|
| sin−1(k) | −2π≤θ≤2π |
| cos−1(k) | 0≤θ≤π |
| tan−1(k) | −2π<θ<2π |
But this is only one of the solutions. To find the rest, you use the CAST diagram.
CAST tells you which trig functions are positive in each quadrant: All in the 1st, Sine in the 2nd, Tangent in the 3rd, Cosine in the 4th. You met this in the Trig Identities lesson.
To find all solutions: get the reference angle α from your calculator, then use CAST to identify which quadrants give solutions. The angles are: 1st: α, 2nd: π−α, 3rd: π+α, 4th: 2π−α.
Solve cosθ=−21 for 0≤θ≤2π.
Reference angle: cos−1(21)=3π
Cosine is negative in the 2nd and 3rd quadrants:
θ=π−3π=32πorθ=π+3π=34π
Always check: cos32π=−21 ✓ and cos34π=−21 ✓
Solve tanθ=3 for 0°≤θ≤360°.
Reference angle: tan−1(3)=60°
Tangent is positive in the 1st and 3rd quadrants:
θ=60°orθ=180°+60°=240°
Answer at least 3 of 3 correctly to complete this section.
When solving sin(nθ)=k or sin(nθ+α)=k, you must extend the range for the substituted variable.
Method:
Solve sin2θ=23 for 0≤θ≤2π.
Let u=2θ. Since θ∈[0,2π], we have u∈[0,4π].
Solve sinu=23. Reference angle: 3π.
Sine is positive in the 1st and 2nd quadrants. In [0,4π]:
u=3π,π−3π=32π,2π+3π=37π,3π−3π=38π
Divide by 2:
θ=6π,3π,67π,34π
Critical check: There should be 4 solutions (double the usual 2, because the "2θ" doubles the range).
Solve cos(2θ+6π)=21 for 0≤θ≤π.
Let u=2θ+6π.
When θ=0: u=6π. When θ=π: u=2π+6π=613π.
So u∈[6π,613π].
Solve cosu=21. Reference angle: 3π.
In the full range [0,2π], cosu=21 gives u=3π or u=35π.
Adding 2π: u=37π — but 37π>613π, so this is outside our range.
Valid values in [6π,613π]: u=3π and u=35π.
Now solve 2θ+6π=3π: θ=21(3π−6π)=12π
And 2θ+6π=35π: θ=21(35π−6π)=21×23π=43π
Solutions: θ=12π or θ=43π.
Answer at least 3 of 3 correctly to complete this section.
When an equation involves sin2θ (or cos2θ) and sinθ (or cosθ), treat it as a quadratic. If it involves both sin and cos, use an identity to convert to a single function first.
Common identity substitutions:
Never divide by sinθ or cosθ unless you are certain they are not zero in your range. Dividing loses solutions where that function equals zero.
Solve 2sin2θ+sinθ−1=0 for 0≤θ≤2π.
Let s=sinθ:
2s2+s−1=0
Factorise: (2s−1)(s+1)=0
s=21ors=−1
When sinθ=21: Reference angle =6π.
θ=6πorθ=π−6π=65π
When sinθ=−1:
θ=23π
Solutions: θ=6π,65π,23π.
Solve 2cos2θ+3sinθ−3=0 for 0°≤θ≤360°.
Replace cos2θ=1−sin2θ:
2(1−sin2θ)+3sinθ−3=0
2−2sin2θ+3sinθ−3=0
−2sin2θ+3sinθ−1=0
Multiply by −1: 2sin2θ−3sinθ+1=0
Factorise: (2sinθ−1)(sinθ−1)=0
sinθ=21orsinθ=1
θ=30°,150°,90°
Solve sinθcosθ=sinθ for 0≤θ≤2π.
Wrong approach: Divide both sides by sinθ to get cosθ=1, giving θ=0 or θ=2π.
This misses all solutions where sinθ=0.
Correct approach: Rearrange and factorise.
sinθcosθ−sinθ=0
sinθ(cosθ−1)=0
Either sinθ=0: θ=0,π,2π
Or cosθ=1: θ=0,2π
All solutions: θ=0,π,2π.
Lesson: Factorising keeps all solutions. Dividing can lose them.
Answer at least 3 of 4 correctly to complete this section.
Before solving any trig equation, ask yourself:
Solve 4sin2θ−4cosθ−1=0 for 0≤θ≤2π. Give your answers in exact form. [5 marks]
Step 1: Replace sin2θ=1−cos2θ:
4(1−cos2θ)−4cosθ−1=0
4−4cos2θ−4cosθ−1=0
−4cos2θ−4cosθ+3=0
Multiply by −1: 4cos2θ+4cosθ−3=0
Step 2: Let c=cosθ:
4c2+4c−3=0
(2c+3)(2c−1)=0
c=−23orc=21
Step 3: Since −1≤cosθ≤1, reject c=−23.
Step 4: Solve cosθ=21. Reference angle: 3π.
Cosine is positive in the 1st and 4th quadrants:
θ=3πorθ=2π−3π=35π
Mark scheme note: You would get 1 mark for the identity substitution, 1 for factorising, 1 for rejecting −23, and 2 for both correct solutions. Forgetting to reject or missing the second solution are the most common ways to lose marks.
Solve tan2θ−3=0 for 0°≤θ≤360°.
tan2θ=3
tanθ=±3
When tanθ=3: θ=60° or θ=240°
When tanθ=−3: θ=120° or θ=300°
Solutions: θ=60°,120°,240°,300°.
Tip: tan2θ=k always gives four solutions in [0°,360°] (provided k>0), because ±k each give two solutions via CAST.
Answer at least 3 of 3 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.