4. (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 the previous step, we obtained:

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

Letting ( y = \sin x ), our equation becomes:

$5y^2 + 3y - 2 = 0.$

To solve for ( y ), we can use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where ( a = 5, b = 3, c = -2. ) Thus we have:

$y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 5 \cdot (-2)}}{2 \cdot 5} = \frac{-3 \pm \sqrt{9 + 40}}{10} = \frac{-3 \pm \sqrt{49}}{10} = \frac{-3 \pm 7}{10}.$

Calculating these gives:

1. ( y_1 = \frac{4}{10} = 0.4 )
2. ( y_2 = \frac{-10}{10} = -1.0 )

To find ( x ) for ( y_1 = 0.4 ):

Using ( \sin x = 0.4 ), we find: [ x = \arcsin(0.4) \approx 23.6^\circ ]
The sine function is also positive in the second quadrant: [ x = 180^\circ - 23.6^\circ \approx 156.4^\circ ]
Combining these, we have the solutions:

• ( x \approx 23.6^\circ )
• ( x \approx 156.4^\circ )

For ( y_2 = -1.0 ): [ \sin x = -1 \Rightarrow x = 270^\circ ]

Thus, the complete solutions are:

• ( x \approx 23.6^\circ )
• ( x \approx 156.4^\circ )
• ( x = 270^\circ )