Standard +0.3 This is a straightforward factorization problem requiring grouping terms (3cosθ common to first pair, 2 common to second pair) to get (2tanθ - 1)(3cosθ + 2) = 0, then solving two basic trig equations. Slightly easier than average since the factorization pattern is standard and the range restriction limits solutions to simple cases.
Or similar partial factorisation; condone sign errors.
\((2\tan\theta - 1)(3\cos\theta + 2) [=0]\) leading to \(\tan\theta = \frac{1}{2}\), \(\cos\theta = -\frac{2}{3}\)
M1
OE. At least 2 out of 4 products correct.
\(26.6°\), \(131.8°\)
A1 A1
WWW. Must be 1 d.p. or better. Final A0 if extra solution within the interval. SC B1 No factorisation: Division by \(2\tan\theta - 1\) leading to \(131.8°\) or division by \(3\cos\theta + 2\) leading to \(26.6°\).
Alternative: \(6\cos\theta\!\left(\frac{\sin\theta}{\cos\theta}\right) - 3\cos\theta + 4\!\left(\frac{\sin\theta}{\cos\theta}\right) - 2 [=0]\) leading to \(2\sin\theta(3\cos\theta+2) - \cos\theta(3\cos\theta+2)[=0]\)
M1
Using \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) and reaching a partial factorisation; condone sign errors.
\((2\sin\theta - \cos\theta)(3\cos\theta + 2)[=0]\) leading to \(\tan\theta = \frac{1}{2}\), \(\cos\theta = -\frac{2}{3}\)
M1
At least 2 out of 4 products correct.
\(26.6°\), \(131.8°\)
A1 A1
WWW. Must be 1 d.p. or better. Final A0 if extra solution within the interval. SC B1 as above.
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3\cos\theta(2\tan\theta - 1) + 2(2\tan\theta - 1) [=0]$ | M1 | Or similar partial factorisation; condone sign errors. |
| $(2\tan\theta - 1)(3\cos\theta + 2) [=0]$ leading to $\tan\theta = \frac{1}{2}$, $\cos\theta = -\frac{2}{3}$ | M1 | OE. At least 2 out of 4 products correct. |
| $26.6°$, $131.8°$ | A1 A1 | WWW. Must be 1 d.p. or better. Final A0 if extra solution within the interval. **SC B1** No factorisation: Division by $2\tan\theta - 1$ leading to $131.8°$ or division by $3\cos\theta + 2$ leading to $26.6°$. |
| **Alternative:** $6\cos\theta\!\left(\frac{\sin\theta}{\cos\theta}\right) - 3\cos\theta + 4\!\left(\frac{\sin\theta}{\cos\theta}\right) - 2 [=0]$ leading to $2\sin\theta(3\cos\theta+2) - \cos\theta(3\cos\theta+2)[=0]$ | M1 | Using $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and reaching a partial factorisation; condone sign errors. |
| $(2\sin\theta - \cos\theta)(3\cos\theta + 2)[=0]$ leading to $\tan\theta = \frac{1}{2}$, $\cos\theta = -\frac{2}{3}$ | M1 | At least 2 out of 4 products correct. |
| $26.6°$, $131.8°$ | A1 A1 | WWW. Must be 1 d.p. or better. Final A0 if extra solution within the interval. **SC B1** as above. |
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