8. (a) Show that the equation 4sin²x + 9cosx - 6 = 0 can be written as 4cos²x - 9cosx + 2 = 0 - Edexcel - A-Level Maths Pure - Question 10 - 2009 - Paper 2

Using the equation from part (a):

$4cos²x - 9cosx + 2 = 0$

This is a quadratic equation in terms of $cosx$. Letting $y = cosx$, we rewrite it as:

$4y² - 9y + 2 = 0$

Applying the quadratic formula, $y = \frac{-b \pm \sqrt{b² - 4ac}}{2a}$, with $a = 4$, $b = -9$, and $c = 2$:

$y = \frac{9 \pm \sqrt{(-9)² - 4(4)(2)}}{2(4)}$

Calculating the discriminant:

$(-9)² - 4(4)(2) = 81 - 32 = 49$

Thus,

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

Calculating the two potential solutions:

1. $y = \frac{16}{8} = 2$ (not valid, since $cosx$ must be between -1 and 1)
2. $y = \frac{2}{8} = \frac{1}{4}$

So, $cosx = \frac{1}{4}$.

Now, to find the angles:

Using $x = cos^{-1}(\frac{1}{4})$ which is approximately $75.5°$.

Since $cos$ is positive in the first and fourth quadrants:

• First solution: $x = 75.5°$
• Second solution: $x = 360° - 75.5° = 284.5°$
• Third solution for the next cycle: $x = 360° + 75.5° = 435.5°$
• Fourth solution: $x = 360° + 284.5° = 644.5°$

Now, we also include $720°$ to check:

• $x = 720° + 75.5° = 795.5°$ (not in range)
• $x = 720° + 284.5° = 1004.5°$ (not in range)

Therefore, the valid solutions for $0 ≤ x < 720°$ to 1 decimal place are:

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