8. (a) Show that the equation $$\cos^2 x = 8 \sin^2 x - 6 \sin x$$ can be written in the form $$(3 \sin x - 1)^2 = 2$$ (b) Hence solve, for $0 \leq x < 360^{\circ}$, $$\cos^2 x = 8 \sin^2 x - 6 \sin x$$ giving your answers to 2 decimal places. - Edexcel - A-Level Maths Pure - Question 9 - 2016 - Paper 2

To show that the given equation can be transformed into the desired form, we start with:

$\cos^2 x = 8 \sin^2 x - 6 \sin x$

Using the Pythagorean identity, we know: $\cos^2 x = 1 - \sin^2 x$

Substituting this into the equation gives:

$1 - \sin^2 x = 8 \sin^2 x - 6 \sin x$

Rearranging this yields:

$1 = 9 \sin^2 x - 6 \sin x$

Now, we can rearrange it to:

$9 \sin^2 x - 6 \sin x - 1 = 0$

Next, we apply completing the square:

$9 \left( \sin^2 x - \frac{2}{3} \sin x \right) = 1$

Now, complete the square for the term in parentheses:

$\sin^2 x - \frac{2}{3} \sin x = \left( \sin x - \frac{1}{3} \right)^2 - \frac{1}{9}$

Thus, we have:

$9 \left( \left( \sin x - \frac{1}{3} \right)^2 - \frac{1}{9} \right) = 1$

Simplifying gives:

$9 \left( \sin x - \frac{1}{3} \right)^2 - 1 = 1$

Then, adding 1 to both sides leads to:

$9 \left( \sin x - \frac{1}{3} \right)^2 = 2$

Dividing through by 9 yields:

$\left( 3 \sin x - 1 \right)^2 = 2$