Grade 12 Maths · Trigonometry · Paper 2

Trigonometric Equations Grade 12

Real Trigonometric Equations past paper questions and answers from 2021 to 2025, with each memo split into the ticks examiners give. Try one, mark yourself, and see exactly which skill cost you marks.

past paper questions: 20 past paper questions
skills: 4 skills
videos: 4 videos

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Question 6.1 · March 2025 · Provincial · 4 marks

6.1 Determine the general solution of

cos (x - 4 5^{\circ}) = sin 2 x

(4)

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More Trigonometric Equations Grade 12 questions and answers

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4. Question 6.2.3 · May/June 2025 · Paper 2 · National (DBE) · 5 marks
6.2 Consider the expression: $sin (A - B) - sin (A + B)$
6.2.3 Hence, determine the solution for $sin 4 x - sin 10 x = sin 3 x$ for $x \in [0^{\circ}; 3 0^{\circ}]$ .(5)
Show the memo (3 ticks)Memo
1. ✓
  Substitute the simplified term and factorize $- 2 cos 7 x sin 3 x = sin 3 x$
  $2 cos 7 x sin 3 x + sin 3 x = 0$
  $sin 3 x (2 cos 7 x + 1) = 0$
  2 marks · Take out a common factor
2. ✓
  Set each factor to zero $sin 3 x = 0$ OR $2 cos 7 x + 1 = 0$
  $sin 3 x = 0$ OR $cos 7 x = - \frac{1}{2}$
  1 mark · Solve a quadratic by factorising
3. ✓
  Solve for x in the given intervalFor $sin 3 x = 0$ :
  $3 x = 0^{\circ} + k \cdot 18 0^{\circ}$
  $x = 0^{\circ}$
  
  For $cos 7 x = - \frac{1}{2}$ (reference angle is $6 0^{\circ}$ , cosine is negative in 2nd quadrant):
  $7 x = 18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$
  $x = \frac{12 0 ^{\circ}}{7} \approx 17, 1 4^{\circ}$
  
  Other general solutions will fall outside the $[0^{\circ}; 3 0^{\circ}]$ domain restriction.
  
  $x = 0^{\circ}$ or $x = 17, 1 4^{\circ}$
  2 marks · Find solutions in an interval
5. Question 7.1.2 · March 2024 · Paper 1 · Provincial · 4 marks
7.1 Given: $cos (A - B) = cos A cos B + sin A sin B$
7.1.2 Hence determine the general solution of the equation $sin (2 x + 5 0^{\circ}) - sin 1 5^{\circ} cos 4 8^{\circ} = sin 4 8^{\circ} cos 1 5^{\circ}$ .(4)
Show the memo (4 ticks)Memo
1. ✓
  Rearrange equation $sin (2 x + 5 0^{\circ}) = sin 4 8^{\circ} cos 1 5^{\circ} + sin 1 5^{\circ} cos 4 8^{\circ}$
  1 mark · Solve a linear equation
2. ✓
  Apply sine compound angle identity $sin (2 x + 5 0^{\circ}) = sin (4 8^{\circ} + 1 5^{\circ})$
  $sin (2 x + 5 0^{\circ}) = sin 6 3^{\circ}$
  1 mark · Use a compound angle identity
3. ✓
  Set up both general solution branches $2 x + 5 0^{\circ} = 6 3^{\circ} + k \cdot 36 0^{\circ}$ OR $2 x + 5 0^{\circ} = 18 0^{\circ} - 6 3^{\circ} + k \cdot 36 0^{\circ}$
  $2 x = 1 3^{\circ} + k \cdot 36 0^{\circ}$ OR $2 x = 6 7^{\circ} + k \cdot 36 0^{\circ}$
  1 mark · Write the general solution
4. ✓
  Final general solutions $x = 6, 5^{\circ} + k \cdot 18 0^{\circ}$ OR $x = 33, 5^{\circ} + k \cdot 18 0^{\circ}, k \in Z$
  1 mark · Write the general solution
6. Question 7.1.2 · March 2022 · Provincial · 5 marks
7.1 For each of the following:
7.1.2 Hence, determine, without the use of a calculator, the general solution of $16 cos^{4} x - 16 cos^{2} x + 2 = 1$ .(5)
Show the memo (3 ticks)Memo
1. ✓
  Rewrite equation using identity $16 cos^{4} x - 16 cos^{2} x + 2 = 1$
  $2 (8 cos^{4} x - 8 cos^{2} x + 1) = 1$
  $2 cos 4 x = 1$
  2 marks · Use a double angle identity
2. ✓
  Solve for 4x $cos 4 x = \frac{1}{2}$
  $4 x = 6 0^{\circ} + k \cdot 36 0^{\circ}$ or $4 x = 30 0^{\circ} + k \cdot 36 0^{\circ}$ (or $- 6 0^{\circ}$ )
  1 mark · Write the general solution
3. ✓
  Solve for x $x = 1 5^{\circ} + k \cdot 9 0^{\circ}$ or $x = 7 5^{\circ} + k \cdot 9 0^{\circ}$
  $k \in Z$
  2 marks · Write the general solution
7. Question 4.1.1 · March 2022 · Provincial · 3 marks
4.1 Given $sin 2 4^{\circ} = m$ and $cos 3 5^{\circ} = n$ . Determine the following in terms of $m$ or $n$ .
4.1.1 $tan 6 6^{\circ}$ (3)
Show the memo (2 ticks)Memo
1. ✓
  Relate angles $6 6^{\circ} = 9 0^{\circ} - 2 4^{\circ}$ , so $tan 6 6^{\circ} = cot 2 4^{\circ}$
  Or use diagram: Opposite = $m$ , Hypotenuse = $1$ , Adjacent = $1 - m^{2}$
  1 mark · Use co-ratio identities
2. ✓
  Determine tangent $tan 2 4^{\circ} = \frac{m}{1 - m ^{2}}$
  $tan 6 6^{\circ} = \frac{1}{t a n 2 4 ^{\circ}} = \frac{1 - m ^{2}}{m}$
  2 marks · Find trig ratios from a sketch
8. Question 5.1 · March 2022 · Provincial · 4 marks
A mountain climber wants to determine the height $PQ$ of a mountain. The climber is standing at $R$ on a flat ground. $R$ and $S$ are in the same horizontal plane as the foot of the mountain $Q$ . From $R$ , he measures the following angles: The angle of elevation of the top of the mountain $P$ is $w^{\circ}$ . $P \hat{R} S$ is $x^{\circ}$ . He then walks $z$ metres to point $S$ and measures $R \hat{S} P$ which is $y^{\circ}$ .
5.1 Show that $PQ = \frac{z s i n y \cdot s i n w}{s i n ( x + y )}$ (4)
Show the memo (3 ticks)Memo
1. ✓
  Calculate angle RPS in triangle PRS $R \hat{P} S = 18 0^{\circ} - (x + y)$
  1 mark · Use angle sum of a triangle
2. ✓
  Apply Sine Rule in triangle PRS $\frac{RP}{s i n y} = \frac{z}{s i n ( 18 0 ^{\circ} - ( x + y ))}$
  $RP = \frac{z s i n y}{s i n ( x + y )}$
  2 marks · Use the sine rule
3. ✓
  Calculate PQ in triangle PQR $\frac{PQ}{PR} = sin w$
  $PQ = PR sin w$
  $PQ = \frac{z s i n y \cdot s i n w}{s i n ( x + y )}$
  1 mark · Calculate a side using trig
9. Question 3.1 · March 2022 · Provincial · 6 marks
3.1 Determine the general solution of: $4 sin x + 2 cos 2 x = 2$ (6)
Show the memo (4 ticks)Memo
1. ✓
  Expand double angle $4 sin x + 2 (1 - 2 sin^{2} x) = 2$
  1 mark · Use a double angle identity
2. ✓
  Simplify to quadratic $4 sin x + 2 - 4 sin^{2} x = 2$
  $4 sin x - 4 sin^{2} x = 0$
  $4 sin x (1 - sin x) = 0$
  1 mark · Write a quadratic in standard form
3. ✓
  Solve for sin x $sin x = 0$ or $sin x = 1$
  1 mark · Solve a quadratic by factorising
4. ✓
  Write general solutionsFor $sin x = 0$ : $x = k \cdot 18 0^{\circ}, k \in Z$
  For $sin x = 1$ : $x = 9 0^{\circ} + k \cdot 36 0^{\circ}, k \in Z$
  3 marks · Write the general solution
10. Question 3.1.2 · March 2021 · Provincial · 2 marks
3.1 In the diagram, reflex $T \hat{O} P = α$ and P has coordinates $(- 5; - 12)$ .

Determine the value of each of the following trigonometric ratios WITHOUT using a calculator:
3.1.2 $tan (18 0^{\circ} - α)$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Reduce angle $tan (18 0^{\circ} - α) = - tan α$
  1 mark · Use a reduction formula
2. ✓
  Calculate value $- tan α = - (\frac{y}{x}) = - (\frac{- 12}{- 5}) = - \frac{12}{5}$
  1 mark · Find trig ratios from a sketch

Trigonometric Equations Grade 12 videos

Short lessons from South African teachers first. Videos load only when you tap play, to save your data.

Trig Triangles Grade 10: Find angleKevinmathscience

General and Specific Solution grade 11:Part 1Kevinmathscience

General and Specific Solution grade 11: Part 2Kevinmathscience

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