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Đáp án + Giải thích các bước giải:

$\cot \bigg(\dfrac{x}{3} + \dfrac{pi}{4}\bigg) = 0$

$\Leftrightarrow \cot \bigg(\dfrac{x}{3} + \dfrac{\pi}{4}\bigg) = \cot \dfrac{\pi}{2}$

$\Leftrightarrow \dfrac{x}{3} + \dfrac{\pi}{4} = \dfrac{\pi}{2} + k\pi (k \in \mathbb{Z})$

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

$\Leftrightarrow x = \dfrac{3\pi}{4} + k3\pi (k \in \mathbb{Z})$

Vậy $S = $`{(3\pi)/4 + k3\pi |  k \in ZZ}`

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$3 \tan (20^o - 4x) + \sqrt{3} = 0$

$\Leftrightarrow \tan (20^o - 4x) = -\dfrac{\sqrt{3}}{3}$

$\Leftrightarrow \tan (20^o - 4x) = \tan 150^o$

$\Leftrightarrow 20^o - 4x = 150^o + k180^o (k \in \mathbb{Z})$

$\Leftrightarrow -4x = 130^o + k180^o (k \in \mathbb{Z})$

$\Leftrightarrow x = -32,5^o + k45^o (k \in \mathbb{Z})$

Vậy $S = $`{-32,5^o + k45^o | k \in ZZ}`

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$\cot (x + 3) - \sqrt{3} = 0$

$\Leftrightarrow \cot (x + 3) = \sqrt{3}$

$\Leftrightarrow \cot (x + 3)= \cot \dfrac{\pi}{6}$

$\Leftrightarrow x + 3 = \dfrac{\pi}{6} + k\pi (k \in \mathbb{Z})$

$\Leftrightarrow x = \dfrac{\pi}{6} - 3 + k\pi (k \in \mathbb{Z})$

Vậy $S = $`{\pi/6 - 3 + k\pi | k \in ZZ}`

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$2 \sin \bigg(4x + \dfrac{\pi}{5}\bigg) + \sqrt{3} = 0$

$\Leftrightarrow \sin \bigg(4x + \dfrac{\pi}{5}\bigg) = -\dfrac{\sqrt{3}}{2}$

$\Leftrightarrow \sin \bigg(4x + \dfrac{\pi}{5}\bigg) = \sin \bigg(-\dfrac{\pi}{3}\bigg)$

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

$\Leftrightarrow$ \(\left[ \begin{array}{l}4x =-\dfrac{8\pi}{15} + k2\pi \\4x = \dfrac{17\pi}{15} + k2\pi\end{array} (k \in \mathbb{Z})\right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}x =-\dfrac{2\pi}{15} + \dfrac{k\pi}{2} \\x = \dfrac{17\pi}{60} + \dfrac{k\pi}{2} \end{array} (k \in \mathbb{Z})\right.\) 

Vậy $S = $`{-(2\pi)/15 + (k\pi)/2 | (17\pi)/60 + (k\pi)/2 | k \in ZZ}`

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$\cos \bigg(\dfrac{\pi}{6} - \dfrac{x}{3}\bigg) + 1 = 0$

$\Leftrightarrow \cos \bigg(\dfrac{\pi}{6} - \dfrac{x}{3}\bigg) = -1$

$\Leftrightarrow \dfrac{\pi}{6} - \dfrac{x}{3} = \pi + k2\pi (k \in \mathbb{Z})$

$\Leftrightarrow -\dfrac{x}{3} = \dfrac{5\pi}{6} + k2\pi (k \in \mathbb{Z})$

$\Leftrightarrow x = -\dfrac{5\pi}{2} + k6\pi (k \in \mathbb{Z})$

Vậy $S = $`{-(5\pi)/2 + k6\pi | k \in ZZ}`