| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve shifted trig equation |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C2 trigonometry. Part (a) requires understanding periodicity of tan (routine), part (b) is a basic transformation (recall), and part (c) involves algebraic manipulation using sin²θ + cos²θ = 1 (shown step-by-step) then solving for multiple angles. While it has several parts worth reasonable marks, each component uses standard techniques without requiring novel insight or complex problem-solving, making it slightly easier than the average A-level question. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 49°\) | B1 | 49 as the only solution in the interval \(0° \leq x < 90°\) |
| \(x = 229°\) | B1 (2 marks) | AWRT 229. Not given if any other solution in the interval \(90° \leq x \leq 360°\). Ignore anything outside \(0° \leq x \leq 360°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Translation | B1 | Accept 'translat…' as equivalent. [T or Tr is NOT sufficient] |
| \(\begin{bmatrix} -30° \\ 0 \end{bmatrix}\) | B1 (2 marks) | OE Accept full equivalent to vector in words provided linked to 'translation/move/shift' and correct direction. (0/2 if >1 transformation) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow 5 + \sin^2\theta = 5\cos\theta + 3\cos^2\theta\) | B1 | Correct RHS |
| \(5 + 1 - \cos^2\theta = 5\cos\theta + 3\cos^2\theta\) | M1 | \(\sin^2\theta = 1 - \cos^2\theta\) used to get a quadratic in \(\cos\theta\) |
| \(6 = 5\cos\theta + 4\cos^2\theta\) or \(4\cos^2\theta + 5\cos\theta - 6 = 0\) | A1 | ACF with like terms collected |
| \((4\cos\theta - 3)(\cos\theta + 2) = 0\) | m1 | Correct quadratic and \((4c \pm 3)(c \pm 2)\); or by formula OE PI by 'correct' 2 values for \(\cos\theta\) |
| Since \(\cos\theta \neq -2\), \(\cos\theta = \frac{3}{4}\) | A1 (5 marks) | CSO AG. Must show that the solution \(\cos\theta = -2\) has been considered and rejected |
| Answer | Marks | Guidance |
|---|---|---|
| \(5 + \sin^2 2x = (5 + 3\cos 2x)\cos 2x \Rightarrow \cos 2x = \frac{3}{4}\) | M1 | Using (c)(i) to reach \(\cos 2x = \frac{3}{4}\); or finding at least 3 solutions of \(\cos\theta = \frac{3}{4}\) and dividing them by 2 |
| \(2x = 0.722(7\ldots),\quad 2\pi - 0.722(7\ldots),\quad 2\pi + 0.722(7\ldots),\quad 4\pi - 0.722(7\ldots)\) | m1 | Valid method to find all four 'positions' of solutions |
| \(x = 0.361,\ 2.78,\ 3.50,\ 5.92\) | A1 (3 marks) | CAO Must be these four 3sf values but ignore any values outside the interval \(0 < x < 2\pi\) |
# Question 9:
## Part 9(a):
$x + 30° = 79°,\quad x + 30° = 180° + 79°$
$x = 49°$ | B1 | 49 as the only solution in the interval $0° \leq x < 90°$
$x = 229°$ | B1 (2 marks) | AWRT 229. Not given if any other solution in the interval $90° \leq x \leq 360°$. Ignore anything outside $0° \leq x \leq 360°$
## Part 9(b):
Translation | B1 | Accept 'translat…' as equivalent. [T or Tr is NOT sufficient]
$\begin{bmatrix} -30° \\ 0 \end{bmatrix}$ | B1 (2 marks) | OE Accept full equivalent to vector in words provided linked to 'translation/move/shift' and correct direction. (0/2 if >1 transformation)
## Part 9(c)(i):
$5 + \sin^2\theta = (5 + 3\cos\theta)\cos\theta$
$\Rightarrow 5 + \sin^2\theta = 5\cos\theta + 3\cos^2\theta$ | B1 | Correct RHS
$5 + 1 - \cos^2\theta = 5\cos\theta + 3\cos^2\theta$ | M1 | $\sin^2\theta = 1 - \cos^2\theta$ used to get a quadratic in $\cos\theta$
$6 = 5\cos\theta + 4\cos^2\theta$ or $4\cos^2\theta + 5\cos\theta - 6 = 0$ | A1 | ACF with like terms collected
$(4\cos\theta - 3)(\cos\theta + 2) = 0$ | m1 | Correct quadratic and $(4c \pm 3)(c \pm 2)$; or by formula OE PI by 'correct' 2 values for $\cos\theta$
Since $\cos\theta \neq -2$, $\cos\theta = \frac{3}{4}$ | A1 (5 marks) | CSO AG. Must show that the solution $\cos\theta = -2$ has been considered and rejected
## Part 9(c)(ii):
$5 + \sin^2 2x = (5 + 3\cos 2x)\cos 2x \Rightarrow \cos 2x = \frac{3}{4}$ | M1 | Using (c)(i) to reach $\cos 2x = \frac{3}{4}$; or finding at least 3 solutions of $\cos\theta = \frac{3}{4}$ and dividing them by 2
$2x = 0.722(7\ldots),\quad 2\pi - 0.722(7\ldots),\quad 2\pi + 0.722(7\ldots),\quad 4\pi - 0.722(7\ldots)$ | m1 | Valid method to find all four 'positions' of solutions
$x = 0.361,\ 2.78,\ 3.50,\ 5.92$ | A1 (3 marks) | CAO Must be these four 3sf values but ignore any values outside the interval $0 < x < 2\pi$9
\begin{enumerate}[label=(\alph*)]
\item Write down the two solutions of the equation $\tan \left( x + 30 ^ { \circ } \right) = \tan 79 ^ { \circ }$ in the interval $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.\\
(2 marks)
\item Describe a single geometrical transformation that maps the graph of $y = \tan x$ onto the graph of $y = \tan \left( x + 30 ^ { \circ } \right)$.
\item \begin{enumerate}[label=(\roman*)]
\item Given that $5 + \sin ^ { 2 } \theta = ( 5 + 3 \cos \theta ) \cos \theta$, show that $\cos \theta = \frac { 3 } { 4 }$.
\item Hence solve the equation $5 + \sin ^ { 2 } 2 x = ( 5 + 3 \cos 2 x ) \cos 2 x$ in the interval $0 < x < 2 \pi$, giving your values of $x$ in radians to three significant figures.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2013 Q9 [12]}}