Đáp án + Giải thích các bước giải:

$\textbf{1}\bigg)$

$\sin \bigg(4x - \dfrac{\pi}{3}\bigg) = \dfrac{1}{2}$

$\Leftrightarrow \sin \bigg(4x - \dfrac{\pi}{3}\bigg) = \sin \dfrac{\pi}{6}$

$\Leftrightarrow \left[ \begin{array}{l}4x - \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi \\ 4x - \dfrac{\pi}{3} = \pi - \dfrac{\pi}{6} + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l}4x = \dfrac{\pi}{2} + k2\pi \\ 4x = \dfrac{5\pi}{6} + \dfrac{\pi}{3} + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi}{8} + \dfrac{k\pi}{2} \\ 4x = \dfrac{7\pi}{6} + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi}{8} + \dfrac{k\pi}{2} \\ x = \dfrac{7\pi}{24} + \dfrac{k\pi}{2} \end{array} \right. (k \in \mathbb{Z})$

Xét $x = \dfrac{\pi}{8} + \dfrac{k\pi}{2} \in [-\pi; 2\pi]$, ta có:

$-\pi \le \dfrac{\pi}{8} + \dfrac{k\pi}{2} \le 2\pi$

$\Leftrightarrow -1 \le \dfrac{1}{8} + \dfrac{k}{2} \le 2$

$\Leftrightarrow -\dfrac{9}{8} \le \dfrac{k}{2} \le \dfrac{15}{8}$

$\Leftrightarrow -\dfrac{9}{4} \le k \le \dfrac{15}{4}$

Vì $k \in \mathbb{Z}$ nên $k \in \{-2; -1; 0; 1; 2; 3\}$

$\Rightarrow x \in$`{-(7\pi)/8; -(3\pi)/8; \pi/8; (5\pi)/8; (9\pi)/8; (13\pi)/8}`

Xét $x = \dfrac{7\pi}{24} + \dfrac{k\pi}{2} \in [-\pi; 2\pi]$, ta có:

$-\pi \le \dfrac{7\pi}{24} + \dfrac{k\pi}{2} \le 2\pi$

$\Leftrightarrow -1 \le \dfrac{7}{24} + \dfrac{k}{2} \le 2$

$\Leftrightarrow -\dfrac{31}{24} \le \dfrac{k}{2} \le \dfrac{41}{24}$

$\Leftrightarrow -\dfrac{31}{12} \le k \le \dfrac{41}{12}$

Vì $k \in \mathbb{Z}$ nên $k \in \{-2; -1; 0; 1; 2; 3\}$

$\Rightarrow x \in$`{-(17\pi)/24; -(5\pi)/24; (7\pi)/24; (19\pi)/24; (31\pi)/24; (43\pi)/24}`

Vậy $x \in$`{-(7\pi)/8; -(17\pi)/24; -(3\pi)/8; -(5\pi)/24; \pi/8; (7\pi)/24; (5\pi)/8; (19\pi)/24; (9\pi)/8; (31\pi)/24; (13\pi)/8; (43\pi)/24}`

$\textbf{2}\bigg) \cos (\pi - 3x) = \cos x$

$\Leftrightarrow \left[ \begin{array}{l}\pi - 3x = x + k2\pi \\ \pi - 3x = -x + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l} -x - 3x = -\pi + k2\pi \\ x - 3x = -\pi + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l} -4x = -\pi + k2\pi \\ -2x = -\pi + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi}{4} - \dfrac{k\pi}{2} \\ x = \dfrac{\pi}{2} - k\pi \end{array} \right. (k \in \mathbb{Z})$

Xét $x = \dfrac{\pi}{4} - \dfrac{k\pi}{2} \in [0; 3\pi]$, ta có:

$0 \le \dfrac{\pi}{4} - \dfrac{k\pi}{2} \le 3\pi$

$\Leftrightarrow 0 \le \dfrac{1}{4} - \dfrac{k}{2} \le 3$

$\Leftrightarrow -\dfrac{1}{4} \le -\dfrac{k}{2} \le \dfrac{11}{4}$

$\Leftrightarrow -\dfrac{11}{2} \le k \le \dfrac{1}{2}$

Vì $k \in \mathbb{Z}$ nên $k \in \{-5; -4; -3; -2; -1; 0\}$

$\Rightarrow x \in$`{\pi/4; (3\pi)/4; (5\pi)/4; (7\pi)/4; (9\pi)/4; (11\pi)/4}`

Xét $x = \dfrac{\pi}{2} - k\pi \in [0; 3\pi]$, ta có:

$0 \le \dfrac{\pi}{2} - k\pi \le 3\pi$

$\Leftrightarrow 0 \le \dfrac{1}{2} - k \le 3$

$\Leftrightarrow -\dfrac{1}{2} \le -k \le \dfrac{5}{2}$

$\Leftrightarrow -\dfrac{5}{2} \le k \le \dfrac{1}{2}$

Vì $k \in \mathbb{Z}$ nên $k \in \{-2; -1; 0\}$

$\Rightarrow x \in$`{\pi/2; (3\pi)/2; (5\pi)/2}`

Vậy $x \in $`{\pi/4; \pi/2; (3\pi)/4; (5\pi)/4; (3\pi)/2; (7\pi)/4; (9\pi)/4; (5\pi)/2; (11\pi)/4}`

$\textbf{3}\bigg)$

$\cos \bigg(\pi t + \dfrac{\pi}{6}\bigg) = -\dfrac{1}{2}$

$\Leftrightarrow \cos \bigg(\pi t + \dfrac{\pi}{6}\bigg) = \cos \bigg(\dfrac{2\pi}{3}\bigg)$

$\Leftrightarrow \left[ \begin{array}{l}\pi t + \dfrac{\pi}{6} = \dfrac{2\pi}{3} + k2\pi \\ \pi t + \dfrac{\pi}{6} = -\dfrac{2\pi}{3} + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l}\pi t = \dfrac{\pi}{2} + k2\pi \\ \pi t = -\dfrac{5\pi}{6} + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l}t = \dfrac{1}{2} + 2k \\ t = -\dfrac{5}{6} + 2k \end{array} \right. (k \in \mathbb{Z})$

Xét $t =\dfrac{1}{2} + 2k \in [0; 10]$, ta có:

$0 \le \dfrac{1}{2} + 2k \le 10$

$\Leftrightarrow -\dfrac{1}{2} \le 2k \le \dfrac{19}{2}$

$\Leftrightarrow -\dfrac{1}{4} \le k \le \dfrac{19}{4}$

Vì $k \in \mathbb{Z}$ nên $k \in \{0; 1; 2; 3; 4\}$

$\Rightarrow t \in$`{1/2; 5/2; 9/2; 13/2; 17/2}`

Xét $t =-\dfrac{5}{6} + 2k \in [0; 10]$, ta có:

$0 \le -\dfrac{5}{6} + 2k \le 10$

$\Leftrightarrow \dfrac{5}{6} \le 2k \le \dfrac{65}{6}$

$\Leftrightarrow \dfrac{5}{12} \le k \le \dfrac{65}{12}$

Vì $k \in \mathbb{Z}$ nên $k \in \{1; 2; 3; 4; 5\}$

$\Rightarrow t \in$`{7/6; 19/6; 31/6; 43/6; 55/6}`

Vậy $t \in$`{1/2; 7/6; 5/2; 19/6; 9/2; 31/6; 13/2; 43/6; 17/2; 55/6}`

$\textbf{4}\bigg)$

$\sin \bigg(2x + \dfrac{\pi}{6}\bigg) = \cos x$

$\Leftrightarrow \sin \bigg(2x + \dfrac{\pi}{6}\bigg) = \sin \bigg(\dfrac{\pi}{2} - x\bigg)$

$\Leftrightarrow \left[ \begin{array}{l}2x + \dfrac{\pi}{6} = \dfrac{\pi}{2} - x + k2\pi \\ 2x + \dfrac{\pi}{6} = \pi - \dfrac{\pi}{2} + x + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l}2x + x = \dfrac{\pi}{2} -\dfrac{\pi}{6} + k2\pi \\ 2x - x = \pi - \dfrac{\pi}{2} - \dfrac{\pi}{6} + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l}3x = \dfrac{\pi}{3} + k2\pi \\ x = \dfrac{\pi}{3} + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi}{9} + \dfrac{k2\pi}{3} \\ x = \dfrac{\pi}{3} + k2\pi \end{array} \right. (k \in \mathbb{Z})$

Xét $x = \dfrac{\pi}{9} + \dfrac{k2\pi}{3}$, ta có:

$\dfrac{\pi}{9} + \dfrac{k2\pi}{3} > 0$

$\Leftrightarrow \dfrac{1}{9} + \dfrac{2k}{3} > 0$

$\Leftrightarrow k > -\dfrac{1}{6}$

$\Rightarrow k = 0$ là giá trị $k$ nhỏ nhất thoả mãn

$\Rightarrow x = \dfrac{\pi}{9} + 0 \cdot \dfrac{2\pi}{3} = \dfrac{\pi}{9}$ là nghiệm dương bé nhất của họ nghiệm trên

Xét $x = \dfrac{\pi}{3} + k2\pi$, ta có:

$\dfrac{\pi}{3} + k2\pi > 0$

$\Leftrightarrow \dfrac{1}{3} + 2k > 0$

$\Leftrightarrow k > -\dfrac{1}{6}$

$\Rightarrow k = 0$ là giá trị $k$ nhỏ nhất thoả mãn

$\Rightarrow x = \dfrac{\pi}{3} + 0 \cdot 2\pi = \dfrac{\pi}{3}$ là nghiệm dương bé nhất của họ nghiệm trên

Mà $\dfrac{\pi}{9} < \dfrac{\pi}{3}$

$\Rightarrow x = \dfrac{\pi}{9}$ là nghiệm dương bé nhất của phương trình

Vậy $x = \dfrac{\pi}{9}$

$\textbf{5}\bigg)$

$\tan \bigg(4x - \dfrac{\pi}{4}\bigg) = 1$

$\Leftrightarrow \tan \bigg(4x - \dfrac{\pi}{4} = \tan \dfrac{\pi}{4}$

$\Leftrightarrow 4x - \dfrac{\pi}{4} = \dfrac{\pi}{4} + k\pi (k \in \mathbb{Z})$

$\Leftrightarrow 4x = \dfrac{\pi}{2} + k\pi (k \in \mathbb{Z})$

$\Leftrightarrow x = \dfrac{\pi}{8} + \dfrac{k\pi}{4} (k \in \mathbb{Z})$

Ta có: $x = \dfrac{\pi}{8} + \dfrac{k\pi}{4} < 0$

$\Leftrightarrow \dfrac{1}{8} + \dfrac{k}{4} < 0$

$\Leftrightarrow k < -\dfrac{1}{2}$

$\Rightarrow k = -1$ là giá trị $k$ lớn nhất thoả mãn

$\Rightarrow x = \dfrac{\pi}{8} - \dfrac{\pi}{4} = -\dfrac{\pi}{8}$ là nghiệm âm lớn nhất của phương trình

Vậy $x = -\dfrac{\pi}{8}$

$\textbf{6}\bigg) \cot \bigg(x + \dfrac{\pi}{3}\bigg) = -1$

$\Leftrightarrow \cot \bigg(x + \dfrac{\pi}{3}\bigg) =\cot \dfrac{3\pi}{4}$

$\Leftrightarrow x + \dfrac{\pi}{3} = \dfrac{3\pi}{4} + k\pi (k \in \mathbb{Z})$

$\Leftrightarrow x = \dfrac{5\pi}{12} + k\pi (k \in \mathbb{Z})$

Ta có: $x = \dfrac{5\pi}{12} + k\pi > 0$

$\Leftrightarrow \dfrac{5}{12} + k > 0$

$\Leftrightarrow k > -\dfrac{5}{12}$

$\Rightarrow k = 0$ là giá trị $k$ nhỏ nhất thoả mãn

$\Rightarrow x = \dfrac{5\pi}{12} + 0 \cdot \pi = \dfrac{5\pi}{12}$ là nghiệm dương bé nhất của phương trình

Vậy $x = \dfrac{5\pi}{12}$