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 10 - 2016 - Paper 2

Using the earlier derived equation, we recognize:

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

By taking the square root, we have two cases:

1. Case 1:
   $3 \sin x - 1 = \sqrt{2}$

   Rearranging we get: $\sin x = \frac{1 + \sqrt{2}}{3}$

2. Case 2:
   $3 \sin x - 1 = -\sqrt{2}$

   This gives: $\sin x = \frac{1 - \sqrt{2}}{3}$

Now we solve for $x$. For both cases, we check the valid $x$ values in the interval $0 \leq x < 360^{\circ}$:

• For Case 1:
  Calculate $\sin^{-1}(\frac{1 + \sqrt{2}}{3})$, and find:

  • $x \approx 53.86^{\circ}$
  • $x \approx 126.41^{\circ}$ (using the sine function symmetry).

• For Case 2:
  Since $\frac{1 - \sqrt{2}}{3}$ is negative, check for obtuse angles.

  • Use reference angles to find potential solutions, yielding:
  • $x \approx 352.06^{\circ}$ and $x \approx 187.94^{\circ}$.

Thus, the solutions to the equation in the range are approximately:

• $x \approx 53.86, 126.41, 187.94, 352.06$ degrees.