Show that $4 \sin x + 2 \cos \left( x + \frac{\pi}{6} \right) = a \cos x + b \sin x$, where $a$ and $b$ are exact values. Express $4 \sin x + 2 \cos \left( x + \frac{\pi}{6} \right)$ in the form $k \cos(x - \alpha)$ where $k > 0$ and $0 < \alpha < \frac{\pi}{2}$. Find the value of $k$ and of $\alpha$. State the minimum value of $4 \sin x + 2 \cos \left( x + \frac{\pi}{6} \right)$ and the corresponding value of $x$. Solve the equation $4 \sin x + 2 \cos \left( x + \frac{\pi}{6} \right) = -2$ for $0 < x < 6$.

1. Expand the cosine term using the addition formula

$4 \sin x + 2 \cos \left( x + \frac{\pi}{6} \right) = 4 \sin x + 2 \left( \cos x \cos \frac{\pi}{6} - \sin x \sin \frac{\pi}{6} \right)$

2. Substitute exact values for the trigonometric ratios of $\frac{\pi}{6}$

$= 4 \sin x + 2 \left( \cos x \left(\frac{\sqrt{3}}{2}\right) - \sin x \left(\frac{1}{2}\right) \right)$

3. Simplify to the required form

$= 4 \sin x + \sqrt{3} \cos x - \sin x$
$= \sqrt{3} \cos x + 3 \sin x$
(where $a = \sqrt{3}$ and $b = 3$)

1. Use the result from part (i) to express the given expression in terms of $\cos x$ and $\sin x$

From part (i), expanding $2 \cos \left( x + \frac{\pi}{6} \right)$ gives $2\left(\cos x \cos \frac{\pi}{6} - \sin x \sin \frac{\pi}{6}\right) = \sqrt{3} \cos x - \sin x$.
So, $4 \sin x + 2 \cos \left( x + \frac{\pi}{6} \right) = \sqrt{3} \cos x + 3 \sin x$.

2. Equate to $k \cos(x - \alpha)$ and find $k$

Let $\sqrt{3} \cos x + 3 \sin x = k \cos(x - \alpha) = k \cos x \cos \alpha + k \sin x \sin \alpha$.
Comparing coefficients:
$k \cos \alpha = \sqrt{3}$
$k \sin \alpha = 3$
$k = \sqrt{(\sqrt{3})^2 + 3^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}$

3. Find $\alpha$

$\tan \alpha = \frac{k \sin \alpha}{k \cos \alpha} = \frac{3}{\sqrt{3}} = \sqrt{3}$
Since $0 < \alpha < \frac{\pi}{2}$, $\alpha = \frac{\pi}{3}$.

1. State the minimum value using the amplitude from part (ii)

From part (ii), the expression is $2\sqrt{3} \cos\left(x - \frac{\pi}{3}\right)$.

The minimum value of $\cos\left(x - \frac{\pi}{3}\right)$ is $-1$.

Minimum value $= -2\sqrt{3}$

2. Find the corresponding value of $x$

For the minimum value, $\cos\left(x - \frac{\pi}{3}\right) = -1$

$x - \frac{\pi}{3} = \pi$

$x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$

1. Use the result from part (ii) to rewrite the equation

From part (ii), $4 \sin x + 2 \cos \left( x + \frac{\pi}{6} \right) = 2\sqrt{3} \cos\left(x - \frac{\pi}{3}\right)$.

Substitute this into the equation:
$2\sqrt{3} \cos\left(x - \frac{\pi}{3}\right) = -2$
$\cos\left(x - \frac{\pi}{3}\right) = -\frac{1}{\sqrt{3}}$

2. Find the basic angle and the possible values for $(x - \frac{\pi}{3})$

Basic angle $\alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \approx 0.95532$ rad.

Since $\cos\left(x - \frac{\pi}{3}\right)$ is negative, the angle lies in the 2nd or 3rd quadrant.
The range for $x$ is $0 < x < 6$, so $-\frac{\pi}{3} < x - \frac{\pi}{3} < 6 - \frac{\pi}{3}$ (approx $-1.047 < x - \frac{\pi}{3} < 4.953$).

$x - \frac{\pi}{3} = \pi - 0.95532 \text{ or } \pi + 0.95532$
$x - \frac{\pi}{3} = 2.1863 \text{ or } 4.0969$

3. Solve for $x$

$x = 2.1863 + \frac{\pi}{3} \text{ or } 4.0969 + \frac{\pi}{3}$
$x = 3.2335 \text{ or } 5.1441$
$x \approx 3.23 \text{ or } 5.14 \text{ (to 3 s.f.)}$