2. (a) Show that the equation 5sin x = 1 + 2cos^2 x can be written in the form 2sin^2 x + 5sin x - 3 = 0 (b) Solve, for 0 ≤ x < 360°, 2sin^2 x + 5sin x - 3 = 0 - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 4

To solve the quadratic equation $2 ext{sin}^2 x + 5 ext{sin} x - 3 = 0$, we can use the quadratic formula:

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

Where:

• $a = 2$
• $b = 5$
• $c = -3$

Calculating the discriminant:

$b^2 - 4ac = 5^2 - 4(2)(-3) = 25 + 24 = 49$

Now substituting into the quadratic formula:

$ext{sin} x = \frac{-5 \pm \sqrt{49}}{2 \cdot 2} = \frac{-5 \pm 7}{4}$

This results in two possible solutions:

1. $ext{sin} x = \frac{2}{4} = \frac{1}{2}$
2. $ext{sin} x = \frac{-12}{4} = -3$ (not possible as $ext{sin} x$ must be in the range [-1, 1])

Thus, we only consider $ext{sin} x = \frac{1}{2}$.

The angles for which $ext{sin} x = \frac{1}{2}$ in the interval $[0°, 360°)$ are:

• $x = 30°$
• $x = 150°$

Therefore, the solutions are:

• $x = 30°$
• $x = 150°$.