Standard +0.3 This is a slightly above-average C2 trigonometric equation. Part (a) requires algebraic manipulation using tan x = sin x/cos x and the Pythagorean identity, which is routine but multi-step. Part (b) involves a standard substitution and solving a quadratic, then handling the compound angle and finding solutions in the given range. The question is structured with clear guidance ('show that' and 'hence'), making it more accessible than if students had to devise the approach independently.
12. (a) Show that the equation
$$6 \cos x - 5 \tan x = 0$$
may be expressed in the form
$$6 \sin ^ { 2 } x + 5 \sin x - 6 = 0$$
(b) Hence solve for \(0 \leqslant \theta < 360 ^ { \circ }\)
$$6 \cos \left( 2 \theta - 10 ^ { \circ } \right) - 5 \tan \left( 2 \theta - 10 ^ { \circ } \right) = 0$$
giving your answers to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\(6\cos x - 5\tan x = 6\cos x - \dfrac{5\sin x}{\cos x}\)
M1
Uses \(\tan x = \dfrac{\sin x}{\cos x}\). May be implied by \(6\cos x - 5\tan x = 0 \Rightarrow 6\cos^2 x - 5\sin x = 0\)
\(6\cos^2 x - 5\sin x = 6(1-\sin^2 x) - 5\sin x\)
M1
Uses \(\cos^2 x = 1 - \sin^2 x\)
\(6\sin^2 x + 5\sin x - 6 = 0\) *
A1*
Fully correct proof with no notational errors, missing brackets, missing variables. Allow proof in terms of different variable but final equation must be in terms of \(x\)
Part (b):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
\(6\sin^2 x + 5\sin x - 6 = 0 \Rightarrow \sin x = \ldots\)
M1
Attempt to solve quadratic for \(\sin x\) or \(\sin(2\theta - 10°)\). Allow if quadratic is clear mis-copy
\(\sin x = \dfrac{2}{3}\) or \(\sin(2\theta - 10°) = \dfrac{2}{3}\)
A1
Correct value. Ignore how referenced. Other root can be ignored whether correct or not
Finds arcsin of \(\dfrac{2}{3}\). Must attempt \(\dfrac{\sin^{-1}(\frac{2}{3}) \pm 10}{2}\). Their \(\sin^{-1}\!\left(\dfrac{2}{3}\right)\) must be a value, not just \(\sin^{-1}\!\left(\text{"}\frac{2}{2}\text{"}\right)\). May be implied by sight of \(25.9°\)
Awrt two correct angles
A1
\(\theta = 25.9°,\ 74.1°,\ 205.9°,\ 254.1°\)
A1
All four angles. Ignore answers outside \((0°, 360°)\) but withhold for extra answers in range. Degree symbols not required
## Question 12:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6\cos x - 5\tan x = 6\cos x - \dfrac{5\sin x}{\cos x}$ | M1 | Uses $\tan x = \dfrac{\sin x}{\cos x}$. May be implied by $6\cos x - 5\tan x = 0 \Rightarrow 6\cos^2 x - 5\sin x = 0$ |
| $6\cos^2 x - 5\sin x = 6(1-\sin^2 x) - 5\sin x$ | M1 | Uses $\cos^2 x = 1 - \sin^2 x$ |
| $6\sin^2 x + 5\sin x - 6 = 0$ * | A1* | Fully correct proof with no notational errors, missing brackets, missing variables. Allow proof in terms of different variable but final equation must be in terms of $x$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6\sin^2 x + 5\sin x - 6 = 0 \Rightarrow \sin x = \ldots$ | M1 | Attempt to solve quadratic for $\sin x$ or $\sin(2\theta - 10°)$. Allow if quadratic is clear mis-copy |
| $\sin x = \dfrac{2}{3}$ or $\sin(2\theta - 10°) = \dfrac{2}{3}$ | A1 | Correct value. Ignore how referenced. Other root can be ignored whether correct or not |
| $2\theta - 10° = \sin^{-1}\!\left(\dfrac{2}{3}\right) = \ldots \Rightarrow \theta = \ldots$ | M1 | Finds arcsin of $\dfrac{2}{3}$. Must attempt $\dfrac{\sin^{-1}(\frac{2}{3}) \pm 10}{2}$. Their $\sin^{-1}\!\left(\dfrac{2}{3}\right)$ must be a value, not just $\sin^{-1}\!\left(\text{"}\frac{2}{2}\text{"}\right)$. May be implied by sight of $25.9°$ |
| Awrt two correct angles | A1 | |
| $\theta = 25.9°,\ 74.1°,\ 205.9°,\ 254.1°$ | A1 | All four angles. Ignore answers outside $(0°, 360°)$ but withhold for extra answers in range. Degree symbols not required |
12. (a) Show that the equation
$$6 \cos x - 5 \tan x = 0$$
may be expressed in the form
$$6 \sin ^ { 2 } x + 5 \sin x - 6 = 0$$
(b) Hence solve for $0 \leqslant \theta < 360 ^ { \circ }$
$$6 \cos \left( 2 \theta - 10 ^ { \circ } \right) - 5 \tan \left( 2 \theta - 10 ^ { \circ } \right) = 0$$
giving your answers to one decimal place.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\
\hfill \mbox{\textit{Edexcel C12 2018 Q12 [8]}}