AQA C3 2006 June — Question 3 9 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2006
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeSolve equation using Pythagorean identities
DifficultyModerate -0.3 This is a straightforward multi-part question on reciprocal trig functions and Pythagorean identities. Part (a) is basic recall (sec x = 1/cos x), part (b) is a guided algebraic manipulation using tan²x = sec²x - 1, and part (c) applies the quadratic formula to solve. All steps are standard textbook exercises with no novel problem-solving required, making it slightly easier than average.
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

3
  1. Solve the equation \(\sec x = 5\), giving all the values of \(x\) in the interval \(0 \leqslant x \leqslant 2 \pi\) in radians to two decimal places.
  2. Show that the equation \(\tan ^ { 2 } x = 3 \sec x + 9\) can be written as $$\sec ^ { 2 } x - 3 \sec x - 10 = 0$$
  3. Solve the equation \(\tan ^ { 2 } x = 3 \sec x + 9\), giving all the values of \(x\) in the interval \(0 \leqslant x \leqslant 2 \pi\) in radians to two decimal places.

3(a)
AnswerMarks Guidance
\(\sec x = 5\); \(\cos x = 0.2\); \(x = 1.37, 4.91\) AWRTM1 A1 A1 3 marks total
3(b)
AnswerMarks Guidance
\(\tan^2 x = 3\sec x + 9\); \(\sec^2 x - 1 = 3\sec x + 9\); \(\sec^2 x - 3\sec x - 10 = 0\)M1 A1 for using \(\sec^2 x = 1 + \tan^2 x\) OE; AG; 2 marks total
3(c)
AnswerMarks Guidance
\((\sec x - 5)(\sec x + 2) = 0\); \(\sec x = 5, -2\); \(\cos x = 0.2, -0.5\); \(x = 1.37, 4.91, 2.09, 4.19\)M1 A1 B1F A1 or use of formula (attempt); any 2 correct or ft their 2 answers in (a); all 4 correct, no extras; 4 marks total
### 3(a)
$\sec x = 5$; $\cos x = 0.2$; $x = 1.37, 4.91$ AWRT | M1 A1 A1 | 3 marks total

### 3(b)
$\tan^2 x = 3\sec x + 9$; $\sec^2 x - 1 = 3\sec x + 9$; $\sec^2 x - 3\sec x - 10 = 0$ | M1 A1 | for using $\sec^2 x = 1 + \tan^2 x$ OE; AG; 2 marks total

### 3(c)
$(\sec x - 5)(\sec x + 2) = 0$; $\sec x = 5, -2$; $\cos x = 0.2, -0.5$; $x = 1.37, 4.91, 2.09, 4.19$ | M1 A1 B1F A1 | or use of formula (attempt); any 2 correct or ft their 2 answers in (a); all 4 correct, no extras; 4 marks total
3
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $\sec x = 5$, giving all the values of $x$ in the interval $0 \leqslant x \leqslant 2 \pi$ in radians to two decimal places.
\item Show that the equation $\tan ^ { 2 } x = 3 \sec x + 9$ can be written as

$$\sec ^ { 2 } x - 3 \sec x - 10 = 0$$
\item Solve the equation $\tan ^ { 2 } x = 3 \sec x + 9$, giving all the values of $x$ in the interval $0 \leqslant x \leqslant 2 \pi$ in radians to two decimal places.
\end{enumerate}

\hfill \mbox{\textit{AQA C3 2006 Q3 [9]}}