Edexcel C3 — Question 2 6 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeSolve equation with reciprocal functions
DifficultyStandard +0.3 This question requires knowledge of the Pythagorean identity cot²x + 1 = cosec²x to convert to a quadratic in cosec x, then solving and finding angles in a given interval. It's slightly above average difficulty due to the reciprocal functions and identity manipulation, but follows a standard pattern for C3 trigonometric equations.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals
2. Find, to 2 decimal places, the solutions of the equation $$3 \cot ^ { 2 } x - 4 \operatorname { cosec } x + \operatorname { cosec } ^ { 2 } x = 0$$ in the interval \(0 \leq x \leq 2 \pi\).
AnswerMarks Guidance
\(3(\cos ec^2 x - 1) - 4 \cos ec x + \cos ec^2 x = 0\)M1
\(4 \cos ec^2 x - 4 \cos ec x - 3 = 0\)M1
\((2 \cos ec x + 1)(2 \cos ec x - 3) = 0\)M1
\(\cos ec x = -\frac{1}{2}\) or \(\frac{3}{2}\)A1
\(\sin x = -2\) (no solutions) or \(\frac{2}{3}\)M1
\(x = 0.73, \pi - 0.7297\)M1
\(x = 0.73, 2.41\) (2dp)A2 (6 marks) 
$3(\cos ec^2 x - 1) - 4 \cos ec x + \cos ec^2 x = 0$ | M1
$4 \cos ec^2 x - 4 \cos ec x - 3 = 0$ | M1
$(2 \cos ec x + 1)(2 \cos ec x - 3) = 0$ | M1
$\cos ec x = -\frac{1}{2}$ or $\frac{3}{2}$ | A1
$\sin x = -2$ (no solutions) or $\frac{2}{3}$ | M1
$x = 0.73, \pi - 0.7297$ | M1
$x = 0.73, 2.41$ (2dp) | A2 | (6 marks)
2. Find, to 2 decimal places, the solutions of the equation

$$3 \cot ^ { 2 } x - 4 \operatorname { cosec } x + \operatorname { cosec } ^ { 2 } x = 0$$

in the interval $0 \leq x \leq 2 \pi$.\\

\hfill \mbox{\textit{Edexcel C3  Q2 [6]}}
This paper (8 questions)
View full paper
Q1 5 Q2 6 Q3 8 Q4 9 Q5 10 Q6 11 Q7 12 Q8 14