8. (a) Show that the equation 4sin²x + 9cosx - 6 = 0 can be written as 4cos²x - 9cosx + 2 = 0 - Edexcel - A-Level Maths Pure - Question 9 - 2009 - Paper 2

Using the rewritten form of the equation:

$4 ext{cos}^2 x - 9 ext{cos} x + 2 = 0$

Letting $y = ext{cos} x$, we can substitute:

$4y^2 - 9y + 2 = 0$

Solving this quadratic using the quadratic formula:

$y = \frac{-b \\pm \sqrt{b^2 - 4ac}}{2a} = \frac{9 \\pm \sqrt{(-9)^2 - 4 \cdot 4 \cdot 2}}{2 \cdot 4}$

Calculating the discriminant:

$= \sqrt{81 - 32} = \sqrt{49} = 7$

Now substituting back in:

$y = \frac{9 \pm 7}{8}$

This gives:

1. $y = \frac{16}{8} = 2$ (not valid since $y$ must be in [-1, 1])
2. $y = \frac{2}{8} = \frac{1}{4}$

Thus we have:

$\text{cos} x = \frac{1}{4}$

Now, to find the angles:

1. $x = \text{cos}^{-1}\left(\frac{1}{4}\right)$ gives approximately $75.5°$
2. The second solution in the range is $x = 360° - 75.5° = 284.5°$
3. Continuing, we find for $x + 360°$:
   • $x = 75.5° + 360° = 435.5°$
   • $x = 284.5° + 360° = 644.5°$

Thus, the solutions satisfying $0 ≤ x < 720°$ are:

• $75.5°$
• $284.5°$
• $435.5°$
• $644.5°$

Final answers rounded to one decimal place are:

• $75.5°$
• $284.5°$
• $435.5°$
• $644.5°$