The problem gives a function $f(x) = \cos^2 x + 2\sqrt{3} \cos x \sin x + 3\sin^2 x$. We need to: (9) Transform $f(x)$ into the form $f(x) = a \sin 2x + b \cos 2x + c$, where $a$, $b$, and $c$ are constants. (10) Find the maximum and minimum values of $f(x)$ in the interval $0 \le x \le \pi$.

1. Problem Description

The problem gives a function $f(x) = \cos^2 x + 2\sqrt{3} \cos x \sin x + 3\sin^2 x$. We need to:

(9) Transform $f(x)$ into the form $f(x) = a \sin 2x + b \cos 2x + c$, where $a$, $b$, and $c$ are constants.

(10) Find the maximum and minimum values of $f(x)$ in the interval $0 \le x \le \pi$.

2. Solution Steps

(9)

First, we rewrite the function using the double angle formulas.

$\sin 2x = 2 \sin x \cos x$

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

$\cos^2 x = \frac{1 + \cos 2x}{2}$

$\sin^2 x = \frac{1 - \cos 2x}{2}$

Substitute these formulas into $f(x)$:

$f(x) = \cos^2 x + 2\sqrt{3} \cos x \sin x + 3\sin^2 x$

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

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

$f(x) = \sqrt{3} \sin 2x - \cos 2x + 2$

So, we have $a = \sqrt{3}$, $b = -1$, and $c = 2$. Therefore,

$f(x) = \sqrt{3} \sin 2x - \cos 2x + 2$

(10)

Now, we need to find the maximum and minimum values of $f(x) = \sqrt{3} \sin 2x - \cos 2x + 2$ for $0 \le x \le \pi$.

We can rewrite $\sqrt{3} \sin 2x - \cos 2x$ in the form $R\sin(2x + \alpha)$.

$R = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = \sqrt{4} = 2$.

So, $f(x) = 2\sin(2x + \alpha) + 2$, where $\cos \alpha = \frac{\sqrt{3}}{2}$ and $\sin \alpha = -\frac{1}{2}$. Thus, $\alpha = -\frac{\pi}{6}$.

Then, $f(x) = 2\sin(2x - \frac{\pi}{6}) + 2$.

Since $0 \le x \le \pi$, we have $0 \le 2x \le 2\pi$.

Therefore, $-\frac{\pi}{6} \le 2x - \frac{\pi}{6} \le 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$.

The maximum value of $\sin(2x - \frac{\pi}{6})$ is 1, which occurs when $2x - \frac{\pi}{6} = \frac{\pi}{2}$, so $2x = \frac{\pi}{2} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$, thus $x = \frac{\pi}{3}$.

The maximum value of $f(x)$ is $2(1) + 2 = 4$.

The minimum value of $\sin(2x - \frac{\pi}{6})$ is -1, which occurs when $2x - \frac{\pi}{6} = \frac{3\pi}{2}$, so $2x = \frac{3\pi}{2} + \frac{\pi}{6} = \frac{10\pi}{6} = \frac{5\pi}{3}$, thus $x = \frac{5\pi}{6}$.

The minimum value of $f(x)$ is $2(-1) + 2 = 0$.

3. Final Answer

(9) $f(x) = \sqrt{3} \sin 2x - \cos 2x + 2$

(10) Maximum value is 4 and minimum value is 0.