CAIE P3 2015 June — Question 3 6 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2015
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeDouble angle with reciprocal functions
DifficultyStandard +0.8 This question requires applying the double angle formula for cotangent (cot 2x = (cot²x - 1)/(2cot x)), substituting into the equation, and solving the resulting quadratic in cot x. While it involves multiple steps and manipulation of reciprocal trig functions, it follows a standard pattern for P3-level questions. The restricted domain adds minor complexity but the algebraic manipulation is straightforward once the substitution is made.
Spec1.05o Trigonometric equations: solve in given intervals

3 Solve the equation \(\cot 2 x + \cot x = 3\) for \(0 ^ { \circ } < x < 180 ^ { \circ }\).

AnswerMarks Guidance
Use correct tan \(2A\) and cot \(A\) formulae to form an equation in tan \(x\)M1
Obtain a correct equation in any formA1
Reduce equation to the form \(\tan^2 x + 6\tan x - 3 = 0\), or equivalentA1
Solve a three term quadratic in tan \(x\) for \(x\), as in Q1M1
Obtain answer, e.g. \(24.9°\) (24.896...)A1
Obtain second answer, e.g. \(98.8\) (98.794) and no others in the given intervalA1
[Ignore outside the given interval. Treat answers in radians as a misread.] Radian answers 0.43452, 1.7243A1 6
Use correct tan $2A$ and cot $A$ formulae to form an equation in tan $x$ | M1 |

Obtain a correct equation in any form | A1 |

Reduce equation to the form $\tan^2 x + 6\tan x - 3 = 0$, or equivalent | A1 |

Solve a three term quadratic in tan $x$ for $x$, as in Q1 | M1 |

Obtain answer, e.g. $24.9°$ (24.896...) | A1 |

Obtain second answer, e.g. $98.8$ (98.794) and no others in the given interval | A1 |

[Ignore outside the given interval. Treat answers in radians as a misread.] Radian answers 0.43452, 1.7243 | A1 | 6 |
3 Solve the equation $\cot 2 x + \cot x = 3$ for $0 ^ { \circ } < x < 180 ^ { \circ }$.

\hfill \mbox{\textit{CAIE P3 2015 Q3 [6]}}