Moderate -0.3 Part (i) is straightforward factorization leading to tan θ = -1 and sin θ = 2/5, requiring basic angle finding. Part (ii) requires rearranging to 4sin x cos x = 3sin x, factoring out sin x, and solving cos x = 3/4 (plus x = 0°, 180°), which is slightly more involved but still a standard C2 technique. Overall slightly easier than average due to routine methods and clear structure.
\(4 \sin x \cos x = 3 \sin x \Rightarrow \sin x(4 \cos x - 3) = 0\)
M1
Other possibilities (after squaring): \(\sin^2 x(16 \sin^2 x - 7) = 0\), \((16 \cos^2 x - 9)(\cos^2 x - 1) = 0\)
\(x = 0, 180\) seen
B1, B1
\(x = 41.4, 318.6\) (AWRT: 41, 319)
B1, B1ft
Guidance:
Part (i):
Answer
Marks
Guidance
\(1^{\text{st}}\) B1 for \(-45\) seen (\(\alpha\), where \(
\alpha
< 90\))
\(2^{\text{nd}}\) B1 for 135 seen, or ft (180 \(+\) \(\alpha\)) if \(\alpha\) is negative, or (\(\alpha\) – 180) if \(\alpha\) is positive.
If \(\tan \theta = k\) is obtained from wrong working, \(2^{\text{nd}}\) B1ft is still available.
Answer
Marks
Guidance
\(3^{\text{rd}}\) B1 for awrt 24 (\(\beta\), where \(
\beta
< 90\))
\(4^{\text{th}}\) B1 for awrt 156, or ft (180 \(-\) \(\beta\)) if \(\beta\) is positive, or \(-(180 + \beta)\) if \(\beta\) is negative.
If \(\sin \theta = k\) is obtained from wrong working, \(4^{\text{th}}\) B1ft is still available.
N.B. Losing \(\sin x = 0\) usually gives a maximum of 3 marks M1M0B0B0B1B1
Part (ii):
\(1^{\text{st}}\) M1 for use of \(\tan x = \frac{\sin x}{\cos x}\). Condone \(\frac{3 \cos x}{3 \cos x}\).
\(2^{\text{nd}}\) M1 for correct work leading to 2 factors (may be implied).
\(1^{\text{st}}\) B1 for 0, \(2^{\text{nd}}\) B1 for 180.
Answer
Marks
Guidance
\(3^{\text{rd}}\) B1 for awrt 41 (\(\gamma\), where \(
\gamma
< 180\))
\(4^{\text{th}}\) B1 for awrt 319, or ft (360 \(-\) \(\gamma\)).
If \(\cos \theta = k\) is obtained from wrong working, \(4^{\text{th}}\) B1ft is still available.
Alternative: (squaring both sides)
\(1^{\text{st}}\) M1 for squaring both sides and using a 'quadratic' identity. e.g. \(16 \sin^2 \theta = 9(\sec^2 \theta - 1)\)
\(2^{\text{nd}}\) M1 for reaching a factorised form. e.g. \((16 \cos^2 \theta - 9)(\cos^2 \theta - 1) = 0\)
Then marks are equivalent to the main scheme. Extra solutions, if not rejected, are penalised as in the main scheme.
For both parts of the question:
Extra solutions outside required range: Ignore
Extra solutions inside required range:
For each pair of B marks, the \(2^{\text{nd}}\) B mark is lost if there are two correct values and one or more extra solution(s), e.g. \(\tan \theta = -1 \Rightarrow \theta = -45, -45, 135\) is B1 B0
Answers in radians:
Loses a maximum of 2 B marks in the whole question (to be deducted at the first and second occurrence).
**Part (i)**
$\tan \theta = -1 \Rightarrow \theta = -45, 135$ | B1, B1ft |
$\sin \theta = \frac{2}{3} \Rightarrow \theta = 23.6, 156.4$ (AWRT: 24, 156) | B1, B1ft |
**Part (ii)**
$4 \sin x = \frac{3 \sin x}{\cos x}$ | M1 |
$4 \sin x \cos x = 3 \sin x \Rightarrow \sin x(4 \cos x - 3) = 0$ | M1 |
Other possibilities (after squaring): $\sin^2 x(16 \sin^2 x - 7) = 0$, $(16 \cos^2 x - 9)(\cos^2 x - 1) = 0$ | |
$x = 0, 180$ seen | B1, B1 |
$x = 41.4, 318.6$ (AWRT: 41, 319) | B1, B1ft |
**Guidance:**
**Part (i):**
$1^{\text{st}}$ B1 for $-45$ seen ($\alpha$, where $|\alpha| < 90$)
$2^{\text{nd}}$ B1 for 135 seen, or ft (180 $+$ $\alpha$) if $\alpha$ is negative, or ($\alpha$ – 180) if $\alpha$ is positive.
If $\tan \theta = k$ is obtained from wrong working, $2^{\text{nd}}$ B1ft is still available.
$3^{\text{rd}}$ B1 for awrt 24 ($\beta$, where $|\beta| < 90$)
$4^{\text{th}}$ B1 for awrt 156, or ft (180 $-$ $\beta$) if $\beta$ is positive, or $-(180 + \beta)$ if $\beta$ is negative.
If $\sin \theta = k$ is obtained from wrong working, $4^{\text{th}}$ B1ft is still available.
N.B. Losing $\sin x = 0$ usually gives a maximum of 3 marks M1M0B0B0B1B1
**Part (ii):**
$1^{\text{st}}$ M1 for use of $\tan x = \frac{\sin x}{\cos x}$. Condone $\frac{3 \cos x}{3 \cos x}$.
$2^{\text{nd}}$ M1 for correct work leading to 2 factors (may be implied).
$1^{\text{st}}$ B1 for 0, $2^{\text{nd}}$ B1 for 180.
$3^{\text{rd}}$ B1 for awrt 41 ($\gamma$, where $|\gamma| < 180$)
$4^{\text{th}}$ B1 for awrt 319, or ft (360 $-$ $\gamma$).
If $\cos \theta = k$ is obtained from wrong working, $4^{\text{th}}$ B1ft is still available.
**Alternative:** (squaring both sides)
$1^{\text{st}}$ M1 for squaring both sides and using a 'quadratic' identity. e.g. $16 \sin^2 \theta = 9(\sec^2 \theta - 1)$
$2^{\text{nd}}$ M1 for reaching a factorised form. e.g. $(16 \cos^2 \theta - 9)(\cos^2 \theta - 1) = 0$
Then marks are equivalent to the main scheme. Extra solutions, if not rejected, are penalised as in the main scheme.
For both parts of the question:
**Extra solutions outside required range:** Ignore
**Extra solutions inside required range:**
For each pair of B marks, the $2^{\text{nd}}$ B mark is lost if there are two correct values and one or more extra solution(s), e.g. $\tan \theta = -1 \Rightarrow \theta = -45, -45, 135$ is B1 B0
**Answers in radians:**
Loses a maximum of 2 B marks in the whole question (to be deducted at the first and second occurrence).
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