Standard +0.3 This is a standard C2 trigonometric equation requiring routine algebraic manipulation (converting tan to sin/cos, factoring a quadratic) followed by solving for angles in a restricted domain. The 'show that' part guides students through the algebra, and part (b) is a direct application with double angles. Slightly above average due to the multi-step nature and double angle application, but follows predictable patterns taught in C2.
13. (a) Show that the equation
$$5 \cos x + 1 = \sin x \tan x$$
can be written in the form
$$6 \cos ^ { 2 } x + \cos x - 1 = 0$$
(b) Hence solve, for \(0 \leqslant \theta < 180 ^ { \circ }\)
$$5 \cos 2 \theta + 1 = \sin 2 \theta \tan 2 \theta$$
giving your answers, where appropriate, to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\(5\cos x + 1 = \sin x \tan x \Rightarrow 5\cos x + 1 = \sin x \times \frac{\sin x}{\cos x}\)
M1
Uses \(\tan x = \frac{\sin x}{\cos x}\) in the given equation; may be seen as \(\tan x \cos x = \sin x\)
\(5\cos^2 x + \cos x = \sin^2 x\)
A1
Correct equation (not involving fractions) in both sin and cos
\(5\cos^2 x + \cos x = 1 - \cos^2 x\)
M1
Replaces \(\sin^2 x\) by \(1 - \cos^2 x\) to produce a quadratic in \(\cos x\) only
\(6\cos^2 x + \cos x - 1 = 0\)
A1*
Proceeds correctly to given answer; notation must be consistent, e.g. \(\cos x\) not \(\cos\), \(\sin^2 x\) not \(\sin x^2\)
(4 marks)
Part (b):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
\(6\cos^2 k + \cos k - 1 = 0 \Rightarrow (3\cos k - 1)(2\cos k + 1) = 0\)
M1
Attempts to factorise and solve expression of the form \(6\cos^2 k + \cos k - 1 = 0\); accept use of formula or GC; may solve \(6y^2 + y - 1 = 0\)
\(\Rightarrow \cos k = \frac{1}{3}, -\frac{1}{2}\)
A1
Correct answers for quadratic; accept with \(k = \theta\) or \(2\theta\) or \(x\) or \(y = \frac{1}{3}, -\frac{1}{2}\)
Either \(\cos 2\theta = \frac{1}{3} \Rightarrow 2\theta = 70.53, 289.47\)
M1
Correct order of operations to produce at least one value for \(\theta\) in range \(0 \to 360°\); look for \(\cos 2\theta = \frac{1}{3} \Rightarrow \theta = \frac{\arccos\left(\frac{1}{3}\right)}{2}\) or \(\cos 2\theta = -\frac{1}{2} \Rightarrow \theta = \frac{180 - \arccos\left(\frac{1}{2}\right)}{2}\)
Or \(\cos 2\theta = -\frac{1}{2} \Rightarrow 2\theta = 120, 240\)
A1
Two of \(\theta =\) awrt \(35.3°, 144.7°, 60°, 120°\)
\(\Rightarrow \theta = 35.3°, 144.7°, 60°, 120°\)
M1
Correct order of operations to find at least one other value for \(\theta\) in range \(0 \to 180°\) from principal value; look for \(\cos 2\theta = \frac{1}{3} \Rightarrow \theta = \frac{360 - \arccos\left(\frac{1}{3}\right)}{2}\) or \(\cos 2\theta = -\frac{1}{2} \Rightarrow \theta = \frac{180 + \arccos\left(\frac{1}{2}\right)}{2}\)
A1
All four of \(\theta =\) awrt \(35.3°, 144.7°, 60°, 120°\) with no additional solutions in range; if answers given in radians \(0.615, 2.526, \frac{\pi}{3}(1.047), \frac{2\pi}{3}(2.094)\) withhold final A1
(6 marks) — Total: 10 marks
## Question 13:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5\cos x + 1 = \sin x \tan x \Rightarrow 5\cos x + 1 = \sin x \times \frac{\sin x}{\cos x}$ | M1 | Uses $\tan x = \frac{\sin x}{\cos x}$ in the given equation; may be seen as $\tan x \cos x = \sin x$ |
| $5\cos^2 x + \cos x = \sin^2 x$ | A1 | Correct equation (not involving fractions) in both sin and cos |
| $5\cos^2 x + \cos x = 1 - \cos^2 x$ | M1 | Replaces $\sin^2 x$ by $1 - \cos^2 x$ to produce a quadratic in $\cos x$ only |
| $6\cos^2 x + \cos x - 1 = 0$ | A1* | Proceeds correctly to given answer; notation must be consistent, e.g. $\cos x$ not $\cos$, $\sin^2 x$ not $\sin x^2$ |
**(4 marks)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6\cos^2 k + \cos k - 1 = 0 \Rightarrow (3\cos k - 1)(2\cos k + 1) = 0$ | M1 | Attempts to factorise and solve expression of the form $6\cos^2 k + \cos k - 1 = 0$; accept use of formula or GC; may solve $6y^2 + y - 1 = 0$ |
| $\Rightarrow \cos k = \frac{1}{3}, -\frac{1}{2}$ | A1 | Correct answers for quadratic; accept with $k = \theta$ or $2\theta$ or $x$ or $y = \frac{1}{3}, -\frac{1}{2}$ |
| Either $\cos 2\theta = \frac{1}{3} \Rightarrow 2\theta = 70.53, 289.47$ | M1 | Correct order of operations to produce at least one value for $\theta$ in range $0 \to 360°$; look for $\cos 2\theta = \frac{1}{3} \Rightarrow \theta = \frac{\arccos\left(\frac{1}{3}\right)}{2}$ or $\cos 2\theta = -\frac{1}{2} \Rightarrow \theta = \frac{180 - \arccos\left(\frac{1}{2}\right)}{2}$ |
| Or $\cos 2\theta = -\frac{1}{2} \Rightarrow 2\theta = 120, 240$ | A1 | Two of $\theta =$ awrt $35.3°, 144.7°, 60°, 120°$ |
| $\Rightarrow \theta = 35.3°, 144.7°, 60°, 120°$ | M1 | Correct order of operations to find at least one other value for $\theta$ in range $0 \to 180°$ from principal value; look for $\cos 2\theta = \frac{1}{3} \Rightarrow \theta = \frac{360 - \arccos\left(\frac{1}{3}\right)}{2}$ or $\cos 2\theta = -\frac{1}{2} \Rightarrow \theta = \frac{180 + \arccos\left(\frac{1}{2}\right)}{2}$ |
| | A1 | All four of $\theta =$ awrt $35.3°, 144.7°, 60°, 120°$ with no additional solutions in range; if answers given in radians $0.615, 2.526, \frac{\pi}{3}(1.047), \frac{2\pi}{3}(2.094)$ withhold final A1 |
**(6 marks) — Total: 10 marks**
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13. (a) Show that the equation
$$5 \cos x + 1 = \sin x \tan x$$
can be written in the form
$$6 \cos ^ { 2 } x + \cos x - 1 = 0$$
(b) Hence solve, for $0 \leqslant \theta < 180 ^ { \circ }$
$$5 \cos 2 \theta + 1 = \sin 2 \theta \tan 2 \theta$$
giving your answers, where appropriate, to one decimal place.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\
\hfill \mbox{\textit{Edexcel C12 2017 Q13 [10]}}