1. (a) Show that $$\frac{\sin 2\theta}{1 + \cos 2\theta} = \tan \theta$$ (b) Hence find, for $-180^\circ \leq \theta < 180^\circ$, all the solutions of $$\frac{2\sin 2\theta}{1 + \cos 2\theta} = 1$$ Give your answers to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 3 - 2010 - Paper 5

From the previous step, substituting into the equation:

1. Replace $\sin 2\theta$ and $\cos 2\theta$:

   $\frac{2 \cdot 2 \sin \theta \cos \theta}{1 + (2 \cos^2 \theta - 1)} = 1$.

2. Simplifying:

   $\frac{4 \sin \theta \cos \theta}{2 \cos^2 \theta} = 1$.

3. This simplifies to:

   $2 \tan \theta = 1 \Rightarrow \tan \theta = \frac{1}{2}$.

4. To find solutions for $-180^\circ \leq \theta < 180^\circ$:

   $\theta = \tan^{-1}\left(\frac{1}{2}\right)$ gives approximately $\theta \approx 26.6^\circ$.

5. Considering the periodic nature of the tangent function, the general solutions are:

   $\theta = 26.6^\circ + 180^\circ k$ where k is any integer.

6. For $-180^\circ \leq \theta < 180^\circ$:

   The other solution is $\theta = 26.6^\circ - 180^\circ \approx -153.4^\circ.$

Thus, the solutions to one decimal place are:

• $heta \\approx 26.6^\circ$
• $heta \\approx -153.4^\circ$.