| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: sin²/cos² substitution |
| Difficulty | Moderate -0.3 This is a standard C2 trigonometric equation requiring routine application of the Pythagorean identity sin²x + cos²x = 1 to convert to a quadratic in cos x, then factorizing or using the quadratic formula. Part (a) is trivial recall of tan = sin/cos. The question is slightly easier than average because the conversion is explicitly guided and the quadratic solving is straightforward. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2\sin\theta=7\cos\theta \Rightarrow \frac{\sin\theta}{\cos\theta}=\frac{7}{2}\) | M1 | \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) clearly used |
| \(\tan\theta=\frac{7}{2}\) | A1 | \(\frac{7}{2}\) OE e.g. 3.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(6\sin^2 x=4+\cos x\), \(6(1-\cos^2 x)=4+\cos x\) | M1 | \(\cos^2 x+\sin^2 x=1\) used |
| \(6-6\cos^2 x=4+\cos x \Rightarrow 6\cos^2 x+\cos x-2=0\) | A1 | CSO AG Be convinced |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Uses \(6\cos^2 x+\cos x-2=0\) | M1 | Uses (b)(i) |
| \((3\cos x+2)(2\cos x-1)=0\) | m1, A1 | Correct factorisation or quadratic formula with \(b^2-4ac\) evaluated correctly |
| \(\cos x=-\frac{2}{3}\), \(\cos x=\frac{1}{2}\) | A1 | CSO Both values correct; accept 3sf |
| \(x=132°, 228°, 60°, 300°\) | B2,1,0 | B1 for any 3 of 4 values correct; deduct 1 mark for each extra solution beyond 4 |
# Question 8:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $2\sin\theta=7\cos\theta \Rightarrow \frac{\sin\theta}{\cos\theta}=\frac{7}{2}$ | M1 | $\tan\theta=\frac{\sin\theta}{\cos\theta}$ clearly used |
| $\tan\theta=\frac{7}{2}$ | A1 | $\frac{7}{2}$ OE e.g. 3.5 |
## Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $6\sin^2 x=4+\cos x$, $6(1-\cos^2 x)=4+\cos x$ | M1 | $\cos^2 x+\sin^2 x=1$ used |
| $6-6\cos^2 x=4+\cos x \Rightarrow 6\cos^2 x+\cos x-2=0$ | A1 | CSO AG Be convinced |
## Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Uses $6\cos^2 x+\cos x-2=0$ | M1 | Uses (b)(i) |
| $(3\cos x+2)(2\cos x-1)=0$ | m1, A1 | Correct factorisation or quadratic formula with $b^2-4ac$ evaluated correctly |
| $\cos x=-\frac{2}{3}$, $\cos x=\frac{1}{2}$ | A1 | CSO Both values correct; accept 3sf |
| $x=132°, 228°, 60°, 300°$ | B2,1,0 | B1 for any 3 of 4 values correct; deduct 1 mark for each extra solution beyond 4 |
---8
\begin{enumerate}[label=(\alph*)]
\item Given that $2 \sin \theta = 7 \cos \theta$, find the value of $\tan \theta$.
\item \begin{enumerate}[label=(\roman*)]
\item Use an appropriate identity to show that the equation
$$6 \sin ^ { 2 } x = 4 + \cos x$$
can be written as
$$6 \cos ^ { 2 } x + \cos x - 2 = 0$$
\item Hence solve the equation $6 \sin ^ { 2 } x = 4 + \cos x$ in the interval $0 ^ { \circ } < x < 360 ^ { \circ }$, giving your answers to the nearest degree.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2012 Q8 [10]}}