Show that the equation 3sin²x + 7sin x = cos²x - 4 can be written in the form 4sin²x + 7sin x + 3 = 0 Hence solve, for 0 ≤ x < 360°, 3sin²x + 7sin x = cos²x - 4 giving your answers to 1 decimal place where appropriate. - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 3

From the previous step, we identified the equation:

$4sin²x + 7sin x + 3 = 0$

We can use the quadratic formula:

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

For our equation, where

• $a = 4$,
• $b = 7$,
• $c = 3$,

This gives us:

$sin x = \frac{-7 \pm \sqrt{(7)² - 4(4)(3)}}{2(4)}$

Calculating the discriminant:

$(7)² - 4(4)(3) = 49 - 48 = 1$

Thus,

$sin x = \frac{-7 \pm 1}{8}$

This yields two solutions:

1. $sin x = \frac{-6}{8} = -0.75$
2. $sin x = \frac{-8}{8} = -1$

Now we find the angles:

• For $sin x = -0.75$, the reference angle is: $x = 180° + 48.59°, ext{and}$ $x = 360° - 48.59°$ This gives: $x \approx 228.6° \text{ and } 311.4°$
• For $sin x = -1$: $x = 270°$

Thus, the solutions in the range $0 ≤ x < 360°$ are:

• $x ≈ 228.6°$
• $x ≈ 311.4°$
• $x = 270°$