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

From the equation:

$(1 - 5 \, \text{cos} 2x) \, \text{sin} 2x = 0,$

we can set each factor to zero:

1. First Factor:
   $\text{sin} 2x = 0$
   This gives: $2x = n\pi \implies x = \frac{n\pi}{2}$
   For $0 \leq x \leq 180^\circ$, we take $n = 0, 1, 2$:

   • $x = 0^\circ$
   • $x = 90^\circ$
   • $x = 180^\circ$

2. Second Factor:
   $1 - 5\text{cos} 2x = 0 \\implies \text{cos} 2x = \frac{1}{5}.$
   Therefore, $2x = \cos^{-1}\left(\frac{1}{5}\right)$
   and also: $2x = 360^\circ - \cos^{-1}\left(\frac{1}{5}\right).$
   Calculating: $\cos^{-1}\left(\frac{1}{5}\right) \approx 78.46^\circ.$
   Thus, $2x \approx 78.46^\circ \\implies x \approx 39.23^\circ,$
   and $2x \approx 360 - 78.46 \approx 281.54^\circ \\implies x \approx 140.77^\circ.$

Thus, the complete solutions for $x$ are:

• $x = 0^\circ$
• $x \approx 39.2^\circ$
• $x \approx 140.8^\circ$ (rounded to 1 decimal place).