| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Convert to quadratic in tan |
| Difficulty | Standard +0.3 This is a standard C2 trigonometric equation requiring routine manipulation using tan x = sin x/cos x to convert to quadratic form (already given), then solving a simple quadratic and finding angles. The 'show that' part guides students through the algebraic manipulation, making this slightly easier than average but still requiring multiple steps and careful handling of the domain. |
| 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 |
|---|---|---|
| Answer | Mark | Guidance |
| \(\tan x(\sin x - \cos x) = 6\cos x\) \(\tan x\left(\frac{\sin x}{\cos x} - 1\right) = 6\) \(\tan x(\tan x - 1) = 6\) \(\tan^2 x - \tan x = 6\) | M1 | Use \(\tan x = \frac{\sin x}{\cos x}\) correctly once. Must be used clearly at least once — either explicitly or by writing e.g. 'divide by \(\cos x\)' at side of solution. Allow M1 for any equiv e.g. \(\sin x = \cos x \tan x\). Allow poor notation e.g. writing just tan rather than \(\tan x\) |
| \(\tan^2 x - \tan x - 6 = 0\) AG | A1 | Obtain \(\tan^2 x - \tan x - 6 = 0\). Correct equation in given form including \(= 0\). Correct notation throughout so A0 if e.g. \(\tan x\) rather than \(\tan x\) seen in solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((\tan x - 3)(\tan x + 2) = 0\) \(\tan x = 3,\ \tan x = -2\) | M1 | Attempt to solve quadratic in \(\tan x\). This M mark is just for solving a 3 term quadratic. Condone any substitution used, inc \(x = \tan x\) |
| \(x = \tan^{-1}(3),\ x = \tan^{-1}(-2)\) | M1 | Attempt to solve \(\tan x = k\) at least once. Attempt \(\tan^{-1} k\) at least once. Not dependent on previous mark so M0M1 possible. If going straight from \(\tan x = k\) to \(x = \ldots\), then award M1 only if their angle is consistent with their \(k\) |
| \(x = 71.6°, 252°, 117°, 297°\) | A1 | Obtain two correct solutions. Allow 3sf or better. Must come from a correct method to solve the quadratic. Allow radian equivs i.e. 1.25 / 4.39 / 2.03 / 5.18 |
| A1 | Obtain all 4 correct solutions and no others in range. Must now all be in degrees. Allow 3sf or better. A0 if other incorrect solutions in range \(0° - 360°\) (but ignore any outside this range). SR If no working shown then allow B1 for each correct solution (max of B3 if in radians, or if extra solutions in range) |
# Question 4:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\tan x(\sin x - \cos x) = 6\cos x$ $\tan x\left(\frac{\sin x}{\cos x} - 1\right) = 6$ $\tan x(\tan x - 1) = 6$ $\tan^2 x - \tan x = 6$ | M1 | Use $\tan x = \frac{\sin x}{\cos x}$ correctly once. Must be used clearly at least once — either explicitly or by writing e.g. 'divide by $\cos x$' at side of solution. Allow M1 for any equiv e.g. $\sin x = \cos x \tan x$. Allow poor notation e.g. writing just tan rather than $\tan x$ |
| $\tan^2 x - \tan x - 6 = 0$ **AG** | A1 | Obtain $\tan^2 x - \tan x - 6 = 0$. Correct equation in given form including $= 0$. Correct notation throughout so A0 if e.g. $\tan x$ rather than $\tan x$ seen in solution |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $(\tan x - 3)(\tan x + 2) = 0$ $\tan x = 3,\ \tan x = -2$ | M1 | Attempt to solve quadratic in $\tan x$. This M mark is just for solving a 3 term quadratic. Condone any substitution used, inc $x = \tan x$ |
| $x = \tan^{-1}(3),\ x = \tan^{-1}(-2)$ | M1 | Attempt to solve $\tan x = k$ at least once. Attempt $\tan^{-1} k$ at least once. Not dependent on previous mark so M0M1 possible. If going straight from $\tan x = k$ to $x = \ldots$, then award M1 only if their angle is consistent with their $k$ |
| $x = 71.6°, 252°, 117°, 297°$ | A1 | Obtain two correct solutions. Allow 3sf or better. Must come from a correct method to solve the quadratic. Allow radian equivs i.e. 1.25 / 4.39 / 2.03 / 5.18 |
| | A1 | Obtain all 4 correct solutions and no others in range. Must now all be in degrees. Allow 3sf or better. A0 if other incorrect solutions in range $0° - 360°$ (but ignore any outside this range). **SR** If no working shown then allow **B1** for each correct solution (max of **B3** if in radians, or if extra solutions in range) |
---4 (i) Show that the equation
$$\sin x - \cos x = \frac { 6 \cos x } { \tan x }$$
can be expressed in the form
$$\tan ^ { 2 } x - \tan x - 6 = 0 .$$
(ii) Hence solve the equation $\sin x - \cos x = \frac { 6 \cos x } { \tan x }$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{OCR C2 2014 Q4 [6]}}