Edexcel C3 — Question 2 7 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Marks7
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Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeMultiple angle equations
DifficultyChallenging +1.2 This question requires converting between tan and sec using the identity sec²x = 1 + tan²x, solving a quadratic in sec(2θ), then finding multiple solutions in the extended range 0° ≤ 2θ ≤ 720°. While it involves several steps and careful angle work, it follows a standard procedure for this type of problem with no novel insight required.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
2. Giving your answers to 1 decimal place, solve the equation $$5 \tan ^ { 2 } 2 \theta - 13 \sec 2 \theta = 1 ,$$ for \(\theta\) in the interval \(0 \leq \theta \leq 360 ^ { \circ }\).
\(5(\sec^2 2\theta - 1) - 13\sec 2\theta = 1\)
\(5\sec^2 2\theta - 13\sec 2\theta - 6 = 0\)
AnswerMarks
\((5\sec 2\theta + 2)(\sec 2\theta - 3) = 0\)M1
\(\sec 2\theta = -\frac{2}{5}\) or \(3\)A1
\(\cos 2\theta = -\frac{5}{2}\) (no solutions) or \(\frac{1}{3}\)
\(2\theta = 70.529, 360-70.529, 360+70.529, 720-70.529\)
AnswerMarks Guidance
\(= 70.529, 289.471, 430.529, 649.471\)B1 M1
\(\theta = 35.3°, 144.7°, 215.3°, 324.7°\) (1dp)A2 (7) 
$5(\sec^2 2\theta - 1) - 13\sec 2\theta = 1$

$5\sec^2 2\theta - 13\sec 2\theta - 6 = 0$

$(5\sec 2\theta + 2)(\sec 2\theta - 3) = 0$ | M1

$\sec 2\theta = -\frac{2}{5}$ or $3$ | A1

$\cos 2\theta = -\frac{5}{2}$ (no solutions) or $\frac{1}{3}$

$2\theta = 70.529, 360-70.529, 360+70.529, 720-70.529$
$= 70.529, 289.471, 430.529, 649.471$ | B1 M1

$\theta = 35.3°, 144.7°, 215.3°, 324.7°$ (1dp) | A2 | **(7)**
2. Giving your answers to 1 decimal place, solve the equation

$$5 \tan ^ { 2 } 2 \theta - 13 \sec 2 \theta = 1 ,$$

for $\theta$ in the interval $0 \leq \theta \leq 360 ^ { \circ }$.\\

\hfill \mbox{\textit{Edexcel C3  Q2 [7]}}
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Q1 6 Q2 7 Q3 7 Q4 9 Q5 10 Q6 11 Q7 12 Q8 13