| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve double/multiple angle equation |
| Difficulty | Moderate -0.3 Part (i) is a straightforward double-angle tangent equation requiring basic rearrangement and consideration of the extended range. Part (ii) requires the Pythagorean identity to convert to a quadratic in sin x, then solving—a standard C2 technique. Both are routine exercises with clear methods, slightly easier than average due to limited problem-solving demand. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan 2x = \frac{1}{3}\), \(2x = 18.4°, 198.4°\) | M1 | Attempt correct solution method. Attempt \(\tan^{-1}\left(\frac{1}{3}\right)\) and then halve answer. |
| \(x = 9.22°, 99.2°\) | A1 | Obtain one of \(9.22°\) or \(99.2°\), or better. Allow radian equiv (0.161 or 1.73). |
| Second angle | A1ft | 3 marks Obtain second correct angle. Maximum 2 marks if angles not in degrees. A0 if extra solutions in given range. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3(1 - \sin^2 x) + 2\sin x - 3 = 0\) | M1 | Use \(\cos^2 x = 1 - \sin^2 x\), aef. Must be used correctly, so \(1 - 3\sin^2 x\) is M0. |
| \(3\sin^2 x - 2\sin x = 0\) | A1 | Obtain \(3\sin^2 x - 2\sin x = 0\). Allow aef, but must be simplified (no constant term). |
| \(\sin x(3\sin x - 2) = 0\) | M1 | Attempt to solve equation to find solutions for \(x\). Must be quadratic in \(\sin x\). Candidates need to be solving for \(x\), so need \(\sin^{-1}\) of at least one solution. |
| \(x = 0°, 180°, 41.8°, 138°\) | A1 | Obtain two of \(0°, 180°, 41.8°, 138°\). |
| All four angles | A1 | 5 marks Obtain all four angles. Must all be in degrees with no extra in given range. |
# Question 7:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan 2x = \frac{1}{3}$, $2x = 18.4°, 198.4°$ | M1 | Attempt correct solution method. Attempt $\tan^{-1}\left(\frac{1}{3}\right)$ and then halve answer. |
| $x = 9.22°, 99.2°$ | A1 | Obtain one of $9.22°$ or $99.2°$, or better. Allow radian equiv (0.161 or 1.73). |
| Second angle | A1ft | **3 marks** Obtain second correct angle. Maximum 2 marks if angles not in degrees. A0 if extra solutions in given range. |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3(1 - \sin^2 x) + 2\sin x - 3 = 0$ | M1 | Use $\cos^2 x = 1 - \sin^2 x$, aef. Must be used correctly, so $1 - 3\sin^2 x$ is M0. |
| $3\sin^2 x - 2\sin x = 0$ | A1 | Obtain $3\sin^2 x - 2\sin x = 0$. Allow aef, but must be simplified (no constant term). |
| $\sin x(3\sin x - 2) = 0$ | M1 | Attempt to solve equation to find solutions for $x$. Must be quadratic in $\sin x$. Candidates need to be solving for $x$, so need $\sin^{-1}$ of at least one solution. |
| $x = 0°, 180°, 41.8°, 138°$ | A1 | Obtain two of $0°, 180°, 41.8°, 138°$. |
| All four angles | A1 | **5 marks** Obtain all four angles. Must all be in degrees with no extra in given range. |
---7 Solve each of the following equations for $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.\\
(i) $3 \tan 2 x = 1$\\
(ii) $3 \cos ^ { 2 } x + 2 \sin x - 3 = 0$
\hfill \mbox{\textit{OCR C2 2011 Q7 [8]}}