The image presents a series of trigonometric problems. 1. Transform $\sin x + \cos x$ and $3\sin x - \sqrt{3}\cos x$ into the form $r\sin(x+\alpha)$, where $r>0$ and $-\pi < \alpha \le \pi$.

1. Problem Description

The image presents a series of trigonometric problems.

1. Transform $\sin x + \cos x$ and $3\sin x - \sqrt{3}\cos x$ into the form $r\sin(x+\alpha)$, where $r>0$ and $-\pi < \alpha \le \pi$.

2. Solve the following equations and inequalities for $0 \le x < 2\pi$:

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

* $\cos 2x + \cos x \ge 0$

* $\cos x + \cos 3x = 0$

* $\cos x + \sqrt{3}(\sin x + 1) \ge 0$

3. Evaluate the following expressions:

* $\sin 10^{\circ} - \sin 70^{\circ} + \sin 130^{\circ}$

* $\frac{\sin 55^{\circ} \sin 35^{\circ}}{\cos 80^{\circ} + \cos 40^{\circ}}$

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

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

* Find the maximum and minimum values of $f(x)$ for $0 \le x \le \pi$.

2. Solution Steps

(1) Transform $\sin x + \cos x$ into $r\sin(x+\alpha)$.

$r = \sqrt{1^2 + 1^2} = \sqrt{2}$

$\alpha$ satisfies $\cos \alpha = \frac{1}{\sqrt{2}}$ and $\sin \alpha = \frac{1}{\sqrt{2}}$, so $\alpha = \frac{\pi}{4}$.

Therefore, $\sin x + \cos x = \sqrt{2} \sin(x + \frac{\pi}{4})$.

(2) Transform $3\sin x - \sqrt{3}\cos x$ into $r\sin(x+\alpha)$.

$r = \sqrt{3^2 + (-\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}$

$\alpha$ satisfies $\cos \alpha = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$ and $\sin \alpha = \frac{-\sqrt{3}}{2\sqrt{3}} = -\frac{1}{2}$, so $\alpha = -\frac{\pi}{6}$.

Therefore, $3\sin x - \sqrt{3}\cos x = 2\sqrt{3}\sin(x - \frac{\pi}{6})$.

(3) $\sin 2x = 2\cos x$

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

$2\sin x \cos x - 2\cos x = 0$

$2\cos x (\sin x - 1) = 0$

$\cos x = 0$ or $\sin x = 1$

If $\cos x = 0$, then $x = \frac{\pi}{2}, \frac{3\pi}{2}$.

If $\sin x = 1$, then $x = \frac{\pi}{2}$.

Thus, $x = \frac{\pi}{2}, \frac{3\pi}{2}$.

(4) $\cos 2x + \cos x \ge 0$

$2\cos^2 x - 1 + \cos x \ge 0$

$2\cos^2 x + \cos x - 1 \ge 0$

$(2\cos x - 1)(\cos x + 1) \ge 0$

Since $\cos x + 1 \ge 0$, we need $2\cos x - 1 \ge 0$ or $\cos x + 1 = 0$.

If $\cos x + 1 = 0$, then $\cos x = -1$, so $x = \pi$.

If $2\cos x - 1 \ge 0$, then $\cos x \ge \frac{1}{2}$, so $0 \le x \le \frac{\pi}{3}$ or $\frac{5\pi}{3} \le x < 2\pi$.

Thus, $0 \le x \le \frac{\pi}{3}$, $x = \pi$, or $\frac{5\pi}{3} \le x < 2\pi$.

(5) $\cos x + \cos 3x = 0$

$2\cos(\frac{x+3x}{2})\cos(\frac{x-3x}{2}) = 0$

$2\cos 2x \cos(-x) = 0$

$2\cos 2x \cos x = 0$

$\cos 2x = 0$ or $\cos x = 0$

If $\cos x = 0$, $x = \frac{\pi}{2}, \frac{3\pi}{2}$.

If $\cos 2x = 0$, $2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$, so $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.

Thus, $x = \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}$.

(6) $\cos x + \sqrt{3}(\sin x + 1) \ge 0$

$\cos x + \sqrt{3}\sin x + \sqrt{3} \ge 0$

$2(\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x) + \sqrt{3} \ge 0$

$2\sin(x + \frac{\pi}{6}) + \sqrt{3} \ge 0$

$\sin(x + \frac{\pi}{6}) \ge -\frac{\sqrt{3}}{2}$

Since $0 \le x < 2\pi$, we have $\frac{\pi}{6} \le x + \frac{\pi}{6} < 2\pi + \frac{\pi}{6}$.

$x + \frac{\pi}{6} \in [\frac{\pi}{6}, \frac{5\pi}{3}] \cup [\frac{11\pi}{3}, \frac{13\pi}{6})$

$x + \frac{\pi}{6} \in [\frac{\pi}{6}, \frac{5\pi}{3}] \cup [\frac{17\pi}{6}, 2\pi + \frac{\pi}{6} )$

$x \in [0, \frac{3\pi}{2}] \cup [\frac{8\pi}{3}, 2\pi)$

So $0 \le x \le \frac{3\pi}{2}$ or $\frac{5\pi}{2} \le x < 2\pi$.

(7) $\sin 10^{\circ} - \sin 70^{\circ} + \sin 130^{\circ}$

$\sin 10^{\circ} + (\sin 130^{\circ} - \sin 70^{\circ}) = \sin 10^{\circ} + 2\cos(\frac{130^{\circ}+70^{\circ}}{2})\sin(\frac{130^{\circ}-70^{\circ}}{2})$

$= \sin 10^{\circ} + 2\cos 100^{\circ} \sin 30^{\circ} = \sin 10^{\circ} + 2(-\sin 10^{\circ}) (\frac{1}{2}) = \sin 10^{\circ} - \sin 10^{\circ} = 0$

(8) $\frac{\sin 55^{\circ} \sin 35^{\circ}}{\cos 80^{\circ} + \cos 40^{\circ}}$

$\sin 55^{\circ} \sin 35^{\circ} = \sin 55^{\circ} \cos 55^{\circ} = \frac{1}{2} \sin 110^{\circ} = \frac{1}{2} \sin 70^{\circ}$

$\cos 80^{\circ} + \cos 40^{\circ} = 2\cos(\frac{80^{\circ}+40^{\circ}}{2})\cos(\frac{80^{\circ}-40^{\circ}}{2}) = 2\cos 60^{\circ} \cos 20^{\circ} = 2(\frac{1}{2})\cos 20^{\circ} = \cos 20^{\circ}$

$= \cos 20^{\circ} = \sin 70^{\circ}$

Therefore, $\frac{\sin 55^{\circ} \sin 35^{\circ}}{\cos 80^{\circ} + \cos 40^{\circ}} = \frac{\frac{1}{2}\sin 70^{\circ}}{\sin 70^{\circ}} = \frac{1}{2}$

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

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

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

$2\sqrt{3}\cos x \sin x = \sqrt{3}\sin 2x$

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

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

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

$f(x) = 2(\frac{\sqrt{3}}{2}\sin 2x - \frac{1}{2}\cos 2x) + 2 = 2\sin(2x - \frac{\pi}{6}) + 2$

$0 \le x \le \pi$, so $0 \le 2x \le 2\pi$, $-\frac{\pi}{6} \le 2x - \frac{\pi}{6} \le 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$.

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

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

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

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

3. Final Answer

(1) $\sqrt{2}\sin(x + \frac{\pi}{4})$

(2) $2\sqrt{3}\sin(x - \frac{\pi}{6})$

(3) $x = \frac{\pi}{2}, \frac{3\pi}{2}$

(4) $0 \le x \le \frac{\pi}{3}$, $x = \pi$, or $\frac{5\pi}{3} \le x < 2\pi$

(5) $x = \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}$

(6) $0 \le x \le \frac{3\pi}{2}$

(7) $0$

(8) $\frac{1}{2}$

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

(10) Maximum: 4, Minimum: 0