Solve the equation $4\cos 2A = 3 - 2\sin A$ for $0 \le A \le 2\pi$. It is given that $\text{f}(x) = 2\cos 6x - \frac{1}{2}$ and $\text{g}(x) = 1 - \sin 3x$. State the period of $\text{f}(x)$ and $\text{g}(x)$, in terms of $\pi$. Sketch, on the same axes, the graphs of for $y = \text{f}(x)$ and $y = \text{g}(x)$ for $0 \le x \le \frac{2\pi}{3}$. Explain how the solutions of the equations in part (i) could be used to find the $x$-coordinates of the points of intersection of the graphs of (ii).

1. Express the equation in terms of $\sin A$ using the double angle formula

$4\cos 2A = 3 - 2\sin A$
$4(1 - 2\sin^2 A) = 3 - 2\sin A$
$4 - 8\sin^2 A = 3 - 2\sin A$

2. Rearrange into a standard quadratic equation and solve for $\sin A$

$8\sin^2 A - 2\sin A - 1 = 0$
$(4\sin A + 1)(2\sin A - 1) = 0$
$\sin A = -\frac{1}{4} \quad \text{or} \quad \sin A = \frac{1}{2}$

3. Solve for $A$ when $\sin A = \frac{1}{2}$

For $\sin A = \frac{1}{2}$:
Since $\sin A$ is positive, $A$ is in the 1st and 2nd quadrants.
Basic angle $\alpha = \frac{\pi}{6}$
$A = \frac{\pi}{6}, \pi - \frac{\pi}{6}$
$A = \frac{\pi}{6}, \frac{5\pi}{6}$

4. Solve for $A$ when $\sin A = -\frac{1}{4}$

For $\sin A = -0.25$:
Since $\sin A$ is negative, $A$ is in the 3rd and 4th quadrants.
Basic angle $\alpha = \sin^{-1}(0.25) \approx 0.25268$ rad
$A = \pi + 0.25268, 2\pi - 0.25268$
$A \approx 3.39, 6.03$ (to 3 s.f.)

1. Find the period of f(x)

For $\text{f}(x) = 2\cos 6x - \frac{1}{2}$, the coefficient of $x$ is $6$.
Period of $\text{f}(x) = \frac{2\pi}{6} = \frac{\pi}{3}$

2. Find the period of g(x)

For $\text{g}(x) = 1 - \sin 3x$, the coefficient of $x$ is $3$.
Period of $\text{g}(x) = \frac{2\pi}{3}$

1. Determine the key features of $y = \text{f}(x) = 2\cos 6x - \frac{1}{2}$

Amplitude $= 2$.
Period $= \frac{2\pi}{6} = \frac{\pi}{3}$.
Vertical shift $= -\frac{1}{2}$.
Maximum value $= 2 - \frac{1}{2} = 1.5$.
Minimum value $= -2 - \frac{1}{2} = -2.5$.
For $0 \le x \le \frac{2\pi}{3}$, there are $2$ full cycles.

2. Determine the key features of $y = \text{g}(x) = 1 - \sin 3x$

Amplitude $= 1$.
Period $= \frac{2\pi}{3}$.
Vertical shift $= 1$.
Maximum value $= 1 - (-1) = 2$.
Minimum value $= 1 - 1 = 0$.
For $0 \le x \le \frac{2\pi}{3}$, there is $1$ full cycle of an inverted sine wave.

3. Sketch $y = \text{f}(x)$

Draw a cosine curve starting at $(0, 1.5)$, going down to a minimum of $-2.5$ at $x = \frac{\pi}{6}$, back to $1.5$ at $x = \frac{\pi}{3}$, down to $-2.5$ at $x = \frac{\pi}{2}$, and ending at $1.5$ at $x = \frac{2\pi}{3}$.

4. Sketch $y = \text{g}(x)$ on the same axes

Draw an inverted sine curve starting at $(0, 1)$, going down to a minimum of $0$ at $x = \frac{\pi}{6}$, crossing $y=1$ at $x = \frac{\pi}{3}$, reaching a maximum of $2$ at $x = \frac{\pi}{2}$, and ending at $1$ at $x = \frac{2\pi}{3}$.

1. Equate the two functions to find the intersection condition and simplify

At the points of intersection, $\text{f}(x) = \text{g}(x)$:
$2\cos 6x - \frac{1}{2} = 1 - \sin 3x$
Multiply by 2:
$4\cos 6x - 1 = 2 - 2\sin 3x$
$4\cos 6x = 3 - 2\sin 3x$

2. Relate the simplified equation to part (i) and explain the connection

Comparing $4\cos 6x = 3 - 2\sin 3x$ with $4\cos 2A = 3 - 2\sin A$ from part (i), we can see they are identical if $A = 3x$.
Thus, the $x$-coordinates can be found by taking the solutions for $A$ from part (i) and dividing them by 3 (i.e., $x = \frac{A}{3}$).