6. (a) Show that the equation $$\tan 2x = 5 \sin 2x$$ can be written in the form $$(1 - 5 \cos 2x) \sin 2x = 0$$ (b) Hence solve, for $0 \leq x \leq 180^{\circ}$, $$\tan 2x = 5 \sin 2x$$ giving your answers to 1 decimal place where appropriate - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 3

Now that we have established the equation

$(1 - 5 \cos 2x) \sin 2x = 0$

We can solve for when each factor is equal to zero:

1. First factor: (\sin 2x = 0)

   • This implies: $2x = n\pi$
   • For $n = 0, 1, 2$, we find:
   • When $n = 0$, $2x = 0$ which gives $x = 0$.
   • When $n = 1$, $2x = \pi$ which gives $x = \frac{\pi}{2} \approx 90^{\circ}$.
   • When $n = 2$, $2x = 2\pi$, which gives $x = \pi = 180^{\circ}$.

   Valid solutions are $x = 0^{\circ}, 90^{\circ}, 180^{\circ}$.

2. Second factor: (1 - 5 \cos 2x = 0)

   • This implies: $\cos 2x = \frac{1}{5}$
   • Therefore, $2x = \cos^{-1}(\frac{1}{5})$
   • Using a calculator:
   • $2x \approx 78.46^{\circ}$, so $x \approx 39.2^{\circ}$.
   • And the solution in the range is:
   • $2x = 360^{\circ} - \cos^{-1}(\frac{1}{5}) \approx 281.54^{\circ},$ thus:
   • $x \approx 140.8^{\circ}$.

Summarizing all solutions:

• $x = 0^{\circ}, 39.2^{\circ}, 90^{\circ}, 140.8^{\circ}, 180^{\circ}$.