Moderate -0.3 This is a straightforward application of reciprocal trig functions and double angle equations. Students need to recognize cot 2θ = 3 means tan 2θ = 1/3, find the principal value using inverse tan, then apply the periodicity of tangent to find all solutions in the range 0° ≤ 2θ ≤ 360°. While it involves multiple angles and reciprocal trig, it's a standard procedural question with no conceptual challenges, making it slightly easier than average.
Rearranged to a quadratic \(= 0\) and attempt to solve by formula oe
\(\tan\theta = -6 \pm \sqrt{36+4}/2 = 0.\) or \(-6.1623\)
\(\theta = 9.22°, 99.22°\)
A1
First correct solution
A1
Second correct solution and no others in range (9.22, 99.22 or better); or SC ft A1 for \(90°\) + their first solution; \(-1\) MR if radians used (0.16, 1.73 or better)
[4]
## Question 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cot 2\theta = 3 \Rightarrow \tan 2\theta = \frac{1}{3}$ | M1 | $\tan = \frac{1}{\cot}$ used, soi |
| $2\theta = 18.43°, \quad \theta = 9.22°$ | A1 | For first correct solution (9.22 or better, e.g. 9.217) |
| $2\theta = 198.43°$ | M1 | For method for second solution for $\theta$ |
| $\theta = 99.22°$ | A1 | Second correct solution and no others in range (99.22 or better); or SC ft A1 for $90°$ + their first solution |
| **OR:** $(2\tan\theta)/(1-\tan^2\theta) = \frac{1}{3}$ | M1 | Use of correct double angle formula |
| $\Rightarrow 6\tan\theta = 1 - \tan^2\theta \Rightarrow \tan^2\theta + 6\tan\theta - 1 = 0$ | M1 | Rearranged to a quadratic $= 0$ and attempt to solve by formula oe |
| $\tan\theta = -6 \pm \sqrt{36+4}/2 = 0.$ or $-6.1623$ | | |
| $\theta = 9.22°, 99.22°$ | A1 | First correct solution |
| | A1 | Second correct solution and no others in range (9.22, 99.22 or better); or SC ft A1 for $90°$ + their first solution; $-1$ MR if radians used (0.16, 1.73 or better) |
| **[4]** | | |
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