8. (i) Solve, for $0 \\leq 0 < \\pi$, the equation $$ \sin 3\theta - \sqrt{3} \cos 3\theta = 0 $$ giving your answers in terms of $\pi$ - Edexcel - A-Level Maths Pure - Question 9 - 2015 - Paper 2

From the equation:

$4\sin^2 x + \cos x = 4 - k$

We can use the Pythagorean identity $\sin^2 x = 1 - \cos^2 x$:

$4(1 - \cos^2 x) + \cos x = 4 - k$

This simplifies to:

$4 - 4\cos^2 x + \cos x = 4 - k$

Rearranging gives:

$-4\cos^2 x + \cos x + k = 0$

Now, we can apply the quadratic formula:

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

where $a = -4$, $b = 1$, and $c = k$:

$\cos x = \frac{-1 \pm \sqrt{1^2 - 4(-4)(k)}}{2(-4)} = \frac{-1 \pm \sqrt{1 + 16k}}{-8} = \frac{1 \mp \sqrt{1 + 16k}}{8}$

So:

$\cos x = \frac{1 - \sqrt{1 + 16k}}{8} \, \text{ or } \, \frac{1 + \sqrt{1 + 16k}}{8}$

Substituting $k = 3$ into the equation:

$4\sin^2 x + \cos x = 4 - 3 \implies 4\sin^2 x + \cos x = 1$

Using $\sin^2 x = 1 - \cos^2 x$:

$4(1 - \cos^2 x) + \cos x = 1\implies 4 - 4\cos^2 x + \cos x = 1$

This simplifies to:

$-4\cos^2 x + \cos x + 3 = 0$

Rearranging gives:

$4\cos^2 x - \cos x - 3 = 0$

Again using the quadratic formula:

$\cos x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-3)}}{2(4)}$

This gives:

$\cos x = \frac{1 \pm \sqrt{1 + 48}}{8} = \frac{1 \pm 7}{8}$

Thus:

1. $\cos x = 1\Rightarrow x = 0^\circ$ (discard since range starts at 0)
2. $\cos x = -\frac{3}{4}\Rightarrow x = \cos^{-1}(-\frac{3}{4})$ This corresponds to two angles in $[0, 360^\circ)$:
   • $x_1 = 180^\circ - \cos^{-1}(-\frac{3}{4})$
   • $x_2 = 360^\circ - \cos^{-1}(-\frac{3}{4})$

Calculating these values gives:

$\cos^{-1}(-\frac{3}{4}) \approx 138.59^\circ$, therefore:

• $x_1 \approx 180^\circ - 138.59^\circ \approx 41.41^\circ$
• $x_2 \approx 360^\circ - 138.59^\circ \approx 221.41^\circ$

So, the values of $x$ are approximately:

$x \approx 41.4^\circ, 221.4^\circ$