The Basics
SOHCAHTOA · Right-angled triangle · Six trig functions · Exact values · Graphs
Right-angled triangle & SOHCAHTOA
sin θ
\[\dfrac{\text{opp}}{\text{hyp}}\]
cos θ
\[\dfrac{\text{adj}}{\text{hyp}}\]
tan θ
\[\dfrac{\text{opp}}{\text{adj}}\]
Six trig functions
sin θ
\(\dfrac{\text{opp}}{\text{hyp}}\)
recip → cosec θ
cos θ
\(\dfrac{\text{adj}}{\text{hyp}}\)
recip → sec θ
tan θ
\(\dfrac{\sin\theta}{\cos\theta}\)
recip → cot θ
cosec θ
\(\dfrac{1}{\sin\theta}\)
sec θ
\(\dfrac{1}{\cos\theta}\)
cot θ
\(\dfrac{\cos\theta}{\sin\theta}\)
Exact values
| θ° | θ rad | sin θ | cos θ | tan θ | cosec θ | sec θ | cot θ |
|---|---|---|---|---|---|---|---|
| 0° | \(0\) | \(0\) | \(1\) | \(0\) | — | \(1\) | — |
| 30° | \(\dfrac{\pi}{6}\) | \(\dfrac{1}{2}\) | \(\dfrac{\sqrt{3}}{2}\) | \(\dfrac{1}{\sqrt{3}}\) | \(2\) | \(\dfrac{2\sqrt{3}}{3}\) | \(\sqrt{3}\) |
| 45° | \(\dfrac{\pi}{4}\) | \(\dfrac{1}{\sqrt{2}}\) | \(\dfrac{1}{\sqrt{2}}\) | \(1\) | \(\sqrt{2}\) | \(\sqrt{2}\) | \(1\) |
| 60° | \(\dfrac{\pi}{3}\) | \(\dfrac{\sqrt{3}}{2}\) | \(\dfrac{1}{2}\) | \(\sqrt{3}\) | \(\dfrac{2\sqrt{3}}{3}\) | \(2\) | \(\dfrac{1}{\sqrt{3}}\) |
| 90° | \(\dfrac{\pi}{2}\) | \(1\) | \(0\) | — | \(1\) | — | \(0\) |
y = sin x | period 360° | amp 1
y = cos x | period 360° | amp 1
y = tan x | period 180° | no amp
Graph transformations — \(y = a\cdot f(bx+c)+d\)
a
Vertical stretch. Amplitude = |a|. Negative a flips the graph.
b
Horizontal stretch. Period = \(\frac{360°}{b}\) for sin/cos, \(\frac{180°}{b}\) for tan.
c
Phase shift — slides graph left/right. Careful with direction.
d
Vertical shift — moves the midline. Max = a+d, Min = −a+d.
Interactive graph explorer — drag the sliders
y = 1·sin(1·x + 0°) + 0
Amplitude
1
Period
360°
Max
1
Min
−1
a — amplitude
1
Try negative a — the graph flips
b — period
1
b=2 → period halves. b=0.5 → period doubles
c — phase shift (°)
0°
Positive c shifts graph left
d — vertical shift
0
Moves the midline up or down
Degrees & Radians
deg → rad
\[\theta_{\text{rad}} = \theta° \times \dfrac{\pi}{180}\]
rad → deg
\[\theta° = \theta_{\text{rad}} \times \dfrac{180}{\pi}\]
Key conversions to memorise
| 30° | \(\dfrac{\pi}{6}\) | 90° | \(\dfrac{\pi}{2}\) |
| 45° | \(\dfrac{\pi}{4}\) | 180° | \(\pi\) |
| 60° | \(\dfrac{\pi}{3}\) | 360° | \(2\pi\) |
⚠ CALCULATOR MODE — CHECK BEFORE EVERY EXAM
Wrong calculator mode = wrong answer even if working is perfect. Always check before you start.
How to switch mode — Casio fx-991 (most common)
- Press SHIFT then SETUP (or MODE on older models)
- Look for Angle Unit — usually option 3 or 4
- Select 1: Deg for degrees | 2: Rad for radians
- Check the top of your screen — it should show D or R as confirmation
Sine rule · Cosine rule · Area of triangle
SINE RULE
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\]
Use when you have a side and its opposite angle. Works for any triangle.
⚠ Ambiguous case — when given two sides and a non-included angle, there may be two possible triangles. Always check if a 2nd solution exists.
COSINE RULE
\[a^2 = b^2 + c^2 - 2bc\cos A\]
rearranged to find angle:
\[\cos A = \frac{b^2 + c^2 - a^2}{2bc}\]
Use when you have all 3 sides, or 2 sides and the included angle.
AREA OF TRIANGLE
\[\text{Area} = \frac{1}{2}ab\sin C\]
Use when you know 2 sides and the included angle. Works for any triangle — not just right-angled.
C is the angle between sides a and b.
WHICH RULE TO USE?
Sine
Given a side + opposite angle. Finding missing side or angle.
Cosine
Given 3 sides, or 2 sides + included angle.
Area
Given 2 sides + included angle C.
Trig Identities
The three IGCSE 0606 identities — all derived from Pythagoras
Identity 1 — Pythagorean
\[\sin^2 A + \cos^2 A = 1\]
Comes from Pythagoras on the unit circle. The foundation.
Identity 2
\[1 + \tan^2 A = \sec^2 A\]
Divide identity 1 through by \(\cos^2 A\)
Identity 3
\[1 + \cot^2 A = \cosec^2 A\]
Divide identity 1 through by \(\sin^2 A\)
Deriving identity 2 — live
- Start: \(\sin^2 A + \cos^2 A = 1\)
- Divide every term by \(\cos^2 A\)
- \(\dfrac{\sin^2 A}{\cos^2 A} + \dfrac{\cos^2 A}{\cos^2 A} = \dfrac{1}{\cos^2 A}\)
- Since \(\frac{\sin}{\cos} = \tan\) and \(\frac{1}{\cos} = \sec\):
- \(\tan^2 A + 1 = \sec^2 A\) ✓
Deriving identity 3 — live
- Start: \(\sin^2 A + \cos^2 A = 1\)
- Divide every term by \(\sin^2 A\)
- \(\dfrac{\sin^2 A}{\sin^2 A} + \dfrac{\cos^2 A}{\sin^2 A} = \dfrac{1}{\sin^2 A}\)
- Since \(\frac{\cos}{\sin} = \cot\) and \(\frac{1}{\sin} = \cosec\):
- \(1 + \cot^2 A = \cosec^2 A\) ✓
Key rearrangements to memorise
\(\sin^2 A = 1 - \cos^2 A\)
Rearrange identity 1
\(\cos^2 A = 1 - \sin^2 A\)
Rearrange identity 1
\(\tan^2 A = \sec^2 A - 1\)
Rearrange identity 2
\(\cot^2 A = \cosec^2 A - 1\)
Rearrange identity 3
Solving Trig Equations
CAST diagram · All solutions · Domain adjustment
CAST diagram
Positivity in each quadrant
A (0°–90°)
All positive
S (90°–180°)
Sin only positive
T (180°–270°)
Tan only positive
C (270°–360°)
Cos only positive
General method
1
Isolate one trig function: \(\sin x = k\), \(\cos x = k\), or \(\tan x = k\)
2
Find principal value: \(x_0 = \sin^{-1}(k)\) — always between −90° and 90°
3
Use CAST to identify which quadrants give that sign → find 2nd angle
4
List ALL solutions in the domain — never stop at one answer
5
For \(\sin(2x)\) or \(\cos(x+30°)\): expand the domain first, solve, then convert back
Finding the 2nd angle
sin
\(180° - x_0\)
cos
\(360° - x_0\)
tan
\(180° + x_0\)
rad
Replace 180° → \(\pi\), 360° → \(2\pi\)
Example — \(\sin x = \dfrac{\sqrt{3}}{2},\;\; 0° \leq x \leq 360°\)
1
Principal value: \(x_0 = 60°\)
2
sin positive → quadrants S and A
3
Q1: \(x=60°\) Q2: \(x = 180°-60° = 120°\)
✓
Answer: \(x = 60°,\; 120°\)
Practice Questions
Let students attempt first, then reveal. 20% teaching, 80% doing.
— Easy
Easy
[2] ~2 min
Given that \(\sin\theta = \dfrac{3}{5}\) and \(\theta\) is acute, find the exact values of \(\cos\theta\) and \(\tan\theta\).
\(\cos\theta = \dfrac{4}{5}\qquad \tan\theta = \dfrac{3}{4}\)
Easy
[3] ~3 min
Solve \(\cos x = -\dfrac{1}{2}\) for \(0° \leq x \leq 360°\).
\(x = 120°\) and \(x = 240°\)
Easy
[2] ~2 min
Write down the amplitude and period of \(y = 4\sin(3x) - 2\).
Amplitude \(= 4\) Period \(= \dfrac{360°}{3} = 120°\)
— Medium
Medium
[4] ~4 min
Solve \(2\cos^2 x - \cos x - 1 = 0\) for \(0° \leq x \leq 360°\).
Factor: \((2\cos x + 1)(\cos x - 1) = 0\)
\(\cos x = -\dfrac{1}{2} \Rightarrow x = 120°,\; 240°\)
\(\cos x = 1 \Rightarrow x = 0°,\; 360°\)
Answer: \(x = 0°,\; 120°,\; 240°,\; 360°\)
\(\cos x = -\dfrac{1}{2} \Rightarrow x = 120°,\; 240°\)
\(\cos x = 1 \Rightarrow x = 0°,\; 360°\)
Answer: \(x = 0°,\; 120°,\; 240°,\; 360°\)
Medium
[4] ~5 min
Prove the identity: \(\dfrac{\sec x - \cos x}{\tan x} \equiv \sin x\)
\(\text{LHS} = \dfrac{\dfrac{1}{\cos x} - \cos x}{\dfrac{\sin x}{\cos x}} = \dfrac{\dfrac{1-\cos^2 x}{\cos x}}{\dfrac{\sin x}{\cos x}} = \dfrac{1-\cos^2 x}{\sin x} = \dfrac{\sin^2 x}{\sin x} = \sin x = \text{RHS}\;\checkmark\)
Medium
[3] ~3 min
Given that \(\tan x = 2\), find the exact value of \(\sec^2 x\) and hence \(\cos^2 x\).
\(\sec^2 x = 1 + \tan^2 x = 1 + 4 = 5\)
\(\cos^2 x = \dfrac{1}{\sec^2 x} = \dfrac{1}{5}\)
\(\cos^2 x = \dfrac{1}{\sec^2 x} = \dfrac{1}{5}\)
— Hard (PYP style)
Hard
[5] ~6 min
Solve \(\cosec^2 x - 2\cot x - 4 = 0\) for \(0° \leq x \leq 360°\).
Replace \(\cosec^2 x = 1 + \cot^2 x\):
\(\cot^2 x - 2\cot x - 3 = 0 \Rightarrow (\cot x-3)(\cot x+1)=0\)
\(\cot x = 3 \Rightarrow \tan x = \dfrac{1}{3} \Rightarrow x = 18.4°,\; 198.4°\)
\(\cot x = -1 \Rightarrow \tan x = -1 \Rightarrow x = 135°,\; 315°\)
\(\cot^2 x - 2\cot x - 3 = 0 \Rightarrow (\cot x-3)(\cot x+1)=0\)
\(\cot x = 3 \Rightarrow \tan x = \dfrac{1}{3} \Rightarrow x = 18.4°,\; 198.4°\)
\(\cot x = -1 \Rightarrow \tan x = -1 \Rightarrow x = 135°,\; 315°\)
Hard
[5] ~5 min
Prove that \(\dfrac{1}{1-\sin x} + \dfrac{1}{1+\sin x} \equiv 2\sec^2 x\)
\(\text{LHS} = \dfrac{(1+\sin x)+(1-\sin x)}{(1-\sin x)(1+\sin x)} = \dfrac{2}{1-\sin^2 x} = \dfrac{2}{\cos^2 x} = 2\sec^2 x = \text{RHS}\;\checkmark\)
Hard
[6] ~8 min
Solve \(3\sin(2x - 30°) = 2\) for \(0° \leq x \leq 360°\). Give answers to 1 d.p.
Let \(u = 2x-30°\). Adjusted domain: \(-30° \leq u \leq 690°\)
\(\sin u = \dfrac{2}{3} \Rightarrow u_0 = 41.8°\)
\(u = 41.8° \Rightarrow x = \dfrac{41.8+30}{2} = 35.9°\)
\(u = 138.2° \Rightarrow x = \dfrac{138.2+30}{2} = 84.1°\)
\(u = 401.8° \Rightarrow x = \dfrac{401.8+30}{2} = 215.9°\)
\(u = 498.2° \Rightarrow x = \dfrac{498.2+30}{2} = 264.1°\)
Answer: \(x = 35.9°,\; 84.1°,\; 215.9°,\; 264.1°\)
\(\sin u = \dfrac{2}{3} \Rightarrow u_0 = 41.8°\)
\(u = 41.8° \Rightarrow x = \dfrac{41.8+30}{2} = 35.9°\)
\(u = 138.2° \Rightarrow x = \dfrac{138.2+30}{2} = 84.1°\)
\(u = 401.8° \Rightarrow x = \dfrac{401.8+30}{2} = 215.9°\)
\(u = 498.2° \Rightarrow x = \dfrac{498.2+30}{2} = 264.1°\)
Answer: \(x = 35.9°,\; 84.1°,\; 215.9°,\; 264.1°\)
Hard PYP-Style Questions
10 exam-style questions — strictly 0606 scope · attempt first, reveal working after
Quick reference — CAST & 2nd angle rules
sin — 2nd angle: \(180° - x_0\)
cos — 2nd angle: \(360° - x_0\)
tan — 2nd angle: \(180° + x_0\)
Compound: expand domain first
Q1 ★★★
[5] ~6 min
Solve \(\cosec^2 x - \cot x = 3\) for \(0° \leq x \leq 360°\).
Replace \(\cosec^2 x = 1 + \cot^2 x\):
\(1 + \cot^2 x - \cot x = 3 \Rightarrow \cot^2 x - \cot x - 2 = 0\)
\((\cot x - 2)(\cot x + 1) = 0\)
\(\cot x = 2 \Rightarrow \tan x = \dfrac{1}{2} \Rightarrow x = 26.6°,\; 206.6°\)
\(\cot x = -1 \Rightarrow \tan x = -1 \Rightarrow x = 135°,\; 315°\)
Answer: \(x = 26.6°,\; 135°,\; 206.6°,\; 315°\)
Key: replace cosec² with identity 3 → quadratic in cot
\(1 + \cot^2 x - \cot x = 3 \Rightarrow \cot^2 x - \cot x - 2 = 0\)
\((\cot x - 2)(\cot x + 1) = 0\)
\(\cot x = 2 \Rightarrow \tan x = \dfrac{1}{2} \Rightarrow x = 26.6°,\; 206.6°\)
\(\cot x = -1 \Rightarrow \tan x = -1 \Rightarrow x = 135°,\; 315°\)
Answer: \(x = 26.6°,\; 135°,\; 206.6°,\; 315°\)
Key: replace cosec² with identity 3 → quadratic in cot
Q2 ★★★
[6] ~8 min
Solve \(2\sin(2x + 45°) = \sqrt{3}\) for \(-180° \leq x \leq 180°\), giving answers to 1 d.p.
\(\sin(2x+45°) = \dfrac{\sqrt{3}}{2}\)
Let \(u = 2x+45°\). Expanded domain: \(-315° \leq u \leq 405°\)
Principal value: \(u_0 = 60°\). Sin positive → Q1 & Q2:
\(u = 60°,\; 120°,\; -300°,\; -240°\)
Convert back \(x = \dfrac{u - 45°}{2}\):
Answer: \(x = -172.5°,\; -142.5°,\; 7.5°,\; 37.5°\)
Key: expand domain for u FIRST — easy mark to lose
Let \(u = 2x+45°\). Expanded domain: \(-315° \leq u \leq 405°\)
Principal value: \(u_0 = 60°\). Sin positive → Q1 & Q2:
\(u = 60°,\; 120°,\; -300°,\; -240°\)
Convert back \(x = \dfrac{u - 45°}{2}\):
Answer: \(x = -172.5°,\; -142.5°,\; 7.5°,\; 37.5°\)
Key: expand domain for u FIRST — easy mark to lose
Q3 ★★★
[5] ~6 min
Find all \(x \in [0,\; 2\pi]\) satisfying \(3\sec^2 x - 2\tan x - 8 = 0\).
Replace \(\sec^2 x = 1 + \tan^2 x\):
\(3(1+\tan^2 x) - 2\tan x - 8 = 0 \Rightarrow 3\tan^2 x - 2\tan x - 5 = 0\)
\((3\tan x - 5)(\tan x + 1) = 0\)
\(\tan x = \dfrac{5}{3} \Rightarrow x = 1.030,\; 4.172 \text{ rad}\)
\(\tan x = -1 \Rightarrow x = \dfrac{3\pi}{4},\; \dfrac{7\pi}{4}\)
Answer: \(x = \dfrac{3\pi}{4},\; \dfrac{7\pi}{4},\; 1.03,\; 4.17\)
Watch: radians domain — don't mix degrees in the answer
\(3(1+\tan^2 x) - 2\tan x - 8 = 0 \Rightarrow 3\tan^2 x - 2\tan x - 5 = 0\)
\((3\tan x - 5)(\tan x + 1) = 0\)
\(\tan x = \dfrac{5}{3} \Rightarrow x = 1.030,\; 4.172 \text{ rad}\)
\(\tan x = -1 \Rightarrow x = \dfrac{3\pi}{4},\; \dfrac{7\pi}{4}\)
Answer: \(x = \dfrac{3\pi}{4},\; \dfrac{7\pi}{4},\; 1.03,\; 4.17\)
Watch: radians domain — don't mix degrees in the answer
Q4 ★★★
[6] ~8 min
\(f(x) = a\sin(bx) + c\) has amplitude 3, period 120°, minimum value \(-1\). Find \(a\), \(b\), \(c\), then solve \(f(x) = 2\) for \(0° \leq x \leq 240°\).
\(a = 3\) (amplitude) · \(b = \dfrac{360°}{120°} = 3\) (period) · \(c = -1 + 3 = 2\) (min = c − a)
\(\therefore f(x) = 3\sin(3x) + 2\)
Solve \(3\sin(3x) + 2 = 2 \Rightarrow \sin(3x) = 0\)
Domain: \(0° \leq 3x \leq 720°\), so \(3x = 0°, 180°, 360°, 540°, 720°\)
Answer: \(x = 0°,\; 60°,\; 120°,\; 180°,\; 240°\)
\(\therefore f(x) = 3\sin(3x) + 2\)
Solve \(3\sin(3x) + 2 = 2 \Rightarrow \sin(3x) = 0\)
Domain: \(0° \leq 3x \leq 720°\), so \(3x = 0°, 180°, 360°, 540°, 720°\)
Answer: \(x = 0°,\; 60°,\; 120°,\; 180°,\; 240°\)
Q5 ★★★
[4] ~5 min
Prove the identity: \((\sec x + \tan x)^2 \equiv \dfrac{1 + \sin x}{1 - \sin x}\)
\(\text{LHS} = \left(\dfrac{1}{\cos x} + \dfrac{\sin x}{\cos x}\right)^2 = \left(\dfrac{1+\sin x}{\cos x}\right)^2 = \dfrac{(1+\sin x)^2}{\cos^2 x}\)
Replace \(\cos^2 x = (1-\sin x)(1+\sin x)\):
\(= \dfrac{(1+\sin x)^2}{(1-\sin x)(1+\sin x)} = \dfrac{1+\sin x}{1-\sin x} = \text{RHS} \;\checkmark\)
Key: combine over common denom → factorise cos²x as difference of squares
Replace \(\cos^2 x = (1-\sin x)(1+\sin x)\):
\(= \dfrac{(1+\sin x)^2}{(1-\sin x)(1+\sin x)} = \dfrac{1+\sin x}{1-\sin x} = \text{RHS} \;\checkmark\)
Key: combine over common denom → factorise cos²x as difference of squares
Q6 ★★★
[5] ~6 min
Solve \(|3\sin x - 1| = 2\) for \(0° \leq x \leq 360°\), giving answers to 1 d.p.
Case 1: \(3\sin x - 1 = 2 \Rightarrow \sin x = 1 \Rightarrow x = 90°\)
Case 2: \(3\sin x - 1 = -2 \Rightarrow \sin x = -\dfrac{1}{3}\)
Principal value: \(\sin^{-1}\!\left(\dfrac{1}{3}\right) = 19.47°\). Sin negative → Q3 & Q4:
\(x = 180° + 19.47° = 199.5°\) · \(x = 360° - 19.47° = 340.5°\)
Answer: \(x = 90°,\; 199.5°,\; 340.5°\)
Case 2: \(3\sin x - 1 = -2 \Rightarrow \sin x = -\dfrac{1}{3}\)
Principal value: \(\sin^{-1}\!\left(\dfrac{1}{3}\right) = 19.47°\). Sin negative → Q3 & Q4:
\(x = 180° + 19.47° = 199.5°\) · \(x = 360° - 19.47° = 340.5°\)
Answer: \(x = 90°,\; 199.5°,\; 340.5°\)
Q7 ★★★
[6] ~7 min
The graph of \(y = p + q\sin(rx)\) has maximum 7, minimum 1, and period 180°. Find \(p\), \(q\), \(r\), state the range, and find the smallest positive \(x\) where \(y = 4\).
\(p = \dfrac{7+1}{2} = 4\) · \(q = \dfrac{7-1}{2} = 3\) · \(r = \dfrac{360°}{180°} = 2\)
\(\therefore y = 4 + 3\sin(2x)\)
Range: \(1 \leq y \leq 7\)
Solve \(4 + 3\sin(2x) = 4 \Rightarrow \sin(2x) = 0\)
\(2x = 0°, 180°, \ldots \Rightarrow x = 0°, 90°, \ldots\)
Smallest positive: \(x = 90°\)
Key: midline = (max+min)/2, half-range = (max−min)/2
\(\therefore y = 4 + 3\sin(2x)\)
Range: \(1 \leq y \leq 7\)
Solve \(4 + 3\sin(2x) = 4 \Rightarrow \sin(2x) = 0\)
\(2x = 0°, 180°, \ldots \Rightarrow x = 0°, 90°, \ldots\)
Smallest positive: \(x = 90°\)
Key: midline = (max+min)/2, half-range = (max−min)/2
Q8 ★★★
[4] ~5 min
Prove: \(\dfrac{1}{1-\cos x} - \dfrac{1}{1+\cos x} \equiv 2\cot x\,\cosec x\)
\(\text{LHS} = \dfrac{(1+\cos x)-(1-\cos x)}{(1-\cos x)(1+\cos x)} = \dfrac{2\cos x}{1-\cos^2 x} = \dfrac{2\cos x}{\sin^2 x}\)
\(= 2\cdot\dfrac{\cos x}{\sin x}\cdot\dfrac{1}{\sin x} = 2\cot x\,\cosec x = \text{RHS} \;\checkmark\)
Key: combine fractions → 1−cos²x = sin²x → split into cot × cosec
\(= 2\cdot\dfrac{\cos x}{\sin x}\cdot\dfrac{1}{\sin x} = 2\cot x\,\cosec x = \text{RHS} \;\checkmark\)
Key: combine fractions → 1−cos²x = sin²x → split into cot × cosec
Q9 ★★★
[4] ~5 min
Given \(\cosec\theta - \sin\theta = 3\), find the exact value of \(\cos^2\theta\).
\(\dfrac{1}{\sin\theta} - \sin\theta = 3 \Rightarrow \dfrac{1-\sin^2\theta}{\sin\theta} = 3 \Rightarrow \dfrac{\cos^2\theta}{\sin\theta} = 3\)
So \(\cos^2\theta = 3\sin\theta \;\cdots(1)\)
Using \(\sin^2\theta + \cos^2\theta = 1\):
\(\sin^2\theta + 3\sin\theta = 1 \Rightarrow \sin\theta = \dfrac{-3+\sqrt{13}}{2}\) (positive root)
\(\cos^2\theta = 3\sin\theta = \dfrac{3\sqrt{13}-9}{2}\)
Key: rewrite cosec as 1/sin → use Pythagorean identity to eliminate sin
So \(\cos^2\theta = 3\sin\theta \;\cdots(1)\)
Using \(\sin^2\theta + \cos^2\theta = 1\):
\(\sin^2\theta + 3\sin\theta = 1 \Rightarrow \sin\theta = \dfrac{-3+\sqrt{13}}{2}\) (positive root)
\(\cos^2\theta = 3\sin\theta = \dfrac{3\sqrt{13}-9}{2}\)
Key: rewrite cosec as 1/sin → use Pythagorean identity to eliminate sin
Q10 ★★★
[5] ~6 min
Solve \(4\sin^2 x + \cos x - 3 = 0\) for \(0° \leq x \leq 360°\).
Replace \(\sin^2 x = 1 - \cos^2 x\):
\(4(1-\cos^2 x) + \cos x - 3 = 0 \Rightarrow 4\cos^2 x - \cos x - 1 = 0\)
Quadratic formula: \(\cos x = \dfrac{1 \pm \sqrt{17}}{8}\)
\(\cos x \approx 0.6404 \Rightarrow x = 50.2°,\; 309.8°\)
\(\cos x \approx -0.3904 \Rightarrow x = 113.0°,\; 247.0°\)
Answer: \(x = 50.2°,\; 113.0°,\; 247.0°,\; 309.8°\)
Key: replace sin²x using Pythagorean identity → quadratic in cos x
\(4(1-\cos^2 x) + \cos x - 3 = 0 \Rightarrow 4\cos^2 x - \cos x - 1 = 0\)
Quadratic formula: \(\cos x = \dfrac{1 \pm \sqrt{17}}{8}\)
\(\cos x \approx 0.6404 \Rightarrow x = 50.2°,\; 309.8°\)
\(\cos x \approx -0.3904 \Rightarrow x = 113.0°,\; 247.0°\)
Answer: \(x = 50.2°,\; 113.0°,\; 247.0°,\; 309.8°\)
Key: replace sin²x using Pythagorean identity → quadratic in cos x
Techniques covered
Q1, Q3 Quadratic in cot/tan via identities 2 & 3
Q2, Q4 Domain expansion for compound angles
Q5, Q8 Identity proof — fractions + difference of squares
Q6 Modulus split into two cases
Q7 Reading p, q, r from max/min/period
Q9, Q10 Reciprocal trig + Pythagorean identity chain
PYP Links
Cambridge 0606 past papers and topic-filtered question banks