7. (a) Show that the equation 3sin² x + 7sin x = cos² x - 4 can be written in the form 4sin² x + 7sin x + 3 = 0 (b) Hence, solve, for 0 ≤ x < 360°, 3sin² x + 7sin x = cos² x - 4 giving your answers to 1 decimal place where appropriate. - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 3

From part (a), we have transformed the equation to:

$4sin² x + 7sin x + 3 = 0$

Letting $y = sin x$, we can rewrite this as:

$4y² + 7y + 3 = 0$

We apply the quadratic formula:

$y = \frac{-b \pm \sqrt{b² - 4ac}}{2a}$

Where $a = 4$, $b = 7$, and $c = 3$:

Calculating the discriminant:

$D = 7² - 4 \times 4 \times 3 = 49 - 48 = 1$

Now substituting into the formula:

$y = \frac{-7 \pm \sqrt{1}}{8} = \frac{-7 \pm 1}{8}$

This gives us:

$y_1 = \frac{-6}{8} = -0.75 \\ y_2 = \frac{-8}{8} = -1$

For $y_1 = -0.75$:

$sin x = -0.75$

This occurs in the third and fourth quadrants. Using the inverse sine:

$x = 180° + \arcsin(0.75) \quad \text{and} \quad x = 360° - \arcsin(0.75)$

Calculating:

$x ≈ 180° + 48.59° ≈ 228.59° \\ x ≈ 360° - 48.59° ≈ 311.41°$

For $y_2 = -1$, we find:

ightarrow x = 270°$$ Thus the solutions are: $$x ≈ 228.6°, 311.4°, 270°$$ Presenting the answers to one decimal place, we conclude: