(a) Show that the equation $$5 \, \cos^2 x = 3(1 + \sin x)$$ can be written as $$5 \, \sin^2 x + 3 \, \sin x - 2 = 0.$$ (b) Hence solve, for $0 \leq x < 360^\circ$, the equation $$5 \, \cos^2 x = 3(1 + \sin x),$$ giving your answers to 1 decimal place where appropriate. - Edexcel - A-Level Maths Pure - Question 6 - 2005 - Paper 2

From part (a), we have established that:

$5\sin^2 x + 3\sin x - 2 = 0.$

We can factor this quadratic equation:

$\left(\sin x - 2\right)(\sin x + \frac{1}{5}) = 0.$

Setting each factor to zero gives us:

1. (\sin x - 2 = 0) which has no solution since (\sin x) cannot exceed 1.
2. (\sin x + \frac{1}{5} = 0) or (\sin x = -\frac{1}{5} ).

Finding angles for ( \sin x = -\frac{1}{5} ):

• The reference angle is given by (x = \arcsin(-\frac{1}{5}) \approx -11.54^\circ). Hence we consider the angles in the 3rd and 4th quadrants:

For the third quadrant: $x = 180^\circ + 11.54^\circ \approx 191.54^\circ.$

For the fourth quadrant: $x = 360^\circ - 11.54^\circ \approx 348.46^\circ.$

Thus, the solutions are approximately:

• (x \approx 191.5^\circ)
• (x \approx 348.5^\circ).