Đáp án:

$\begin{array}{l}
a)\tan 9x = 0\\
 \Leftrightarrow \sin 9x = 0\\
 \Leftrightarrow 9x = k\pi \\
 \Leftrightarrow x = \dfrac{{k\pi }}{9}\left( {k \in Z} \right)\\
Vậy\,x = \dfrac{{k\pi }}{9}\left( {k \in Z} \right)\\
b)\tan \left( {3x + \dfrac{\pi }{6}} \right) =  - \dfrac{1}{{\sqrt 3 }}\\
 \Leftrightarrow 3x + \dfrac{\pi }{6} =  - \dfrac{\pi }{6} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow 3x = \dfrac{{ - \pi }}{3} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{{ - \pi }}{9} + \dfrac{{k\pi }}{3}\left( {k \in Z} \right)\\
Vậy\,x = \dfrac{{ - \pi }}{9} + \dfrac{{k\pi }}{3}\left( {k \in Z} \right)\\
c)\tan \left( {2x + \dfrac{\pi }{5}} \right) = \sqrt 3 \\
 \Leftrightarrow 2x + \dfrac{\pi }{5} = \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow 2x = \dfrac{{2\pi }}{{15}} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{\pi }{{15}} + \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
Vậy\,x = \dfrac{\pi }{{15}} + \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
d)\cot 12x = 1\\
 \Leftrightarrow 12x = \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{\pi }{{48}} + \dfrac{{k\pi }}{{12}}\left( {k \in Z} \right)\\
e)\cot 5x =  - \dfrac{1}{{\sqrt 3 }}\\
 \Leftrightarrow 5x = \dfrac{{ - \pi }}{3} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{{ - \pi }}{{15}} + \dfrac{{k\pi }}{5}\left( {k \in Z} \right)\\
Vậy\,x = \dfrac{{ - \pi }}{{15}} + \dfrac{{k\pi }}{5}\left( {k \in Z} \right)\\
f)\cot \left( {3x + {{10}^0}} \right) = \dfrac{{\sqrt 3 }}{3}\\
 \Leftrightarrow 3x + {10^0} = {60^0} + k{.180^0}\left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{{{{50}^0}}}{3} + k{.60^0}\left( {k \in Z} \right)\\
Vậy\,x = \dfrac{{{{50}^0}}}{3} + k{.60^0}\left( {k \in Z} \right)
\end{array}$

Đáp án:

Giải thích các bước giải:

 a) `tan\ 9x=0`

ĐK: `cos\ 9x \ne 0`

`⇔ 9x \ne \frac{\pi}{2}+k\pi\ (k \in \mathbb{Z})`

`⇔ x \ne \frac{\pi}{18}+k\frac{\pi}{9}\ (k \in \mathbb{Z})`

`⇒ 9x = k\pi\ (k \in \mathbb{Z})`

`⇔ x = k\frac{\pi}{9}\ (TM)\ (k \in \mathbb{Z})`

Vậy `S={k\frac{\pi}{9}\ (k \in \mathbb{Z})}`

b) `tan (3x+\frac{\pi}{6})=-\frac{1}{\sqrt{3}}`

ĐK: `cos (3x+\frac{\pi}{6}) \ne 0`

`⇔ 3x+\frac{\pi}{6} \ne \frac{\pi}{2}+k\pi\ (k \in \mathbb{Z})`

`⇔ x \ne \frac{\pi}{9}+k\frac{\pi}{3}\ (k \in \mathbb{Z})`

`⇒ tan (3x+\frac{\pi}{6})=tan (-\frac{\pi}{6})`

`⇔ 3x+\frac{\pi}{6}=-\frac{\pi}{6}+k\pi\ (k \in \mathbb{Z})`

`⇔ x=-\frac{\pi}{9}+k\frac{\pi}{3}\ (TM)\ (k \in \mathbb{Z})`

Vậy `S={-\frac{\pi}{9}+k\frac{\pi}{3}\ (k \in \mathbb{Z})}`

c) `tan (2x+\frac{\pi}{5})=-\sqrt{3}`

ĐK: `cos (2x+\frac{\pi}{5}) \ne 0`

`⇔ 2x+\frac{\pi}{5} \ne \frac{\pi}{2}+k\pi\ (k \in \mathbb{Z})`

`⇔ x \ne \frac{3\pi}{20}+k\frac{\pi}{2}\ (k \in \mathbb{Z})`

`⇒ tan (2x+\frac{\pi}{5})=tan (-\frac{\pi}{3})`

`⇔ 2x+\frac{\pi}{5}=-\frac{\pi}{3}+k\pi\ (k \in \mathbb{Z})`

`⇔ x=-\frac{4\pi}{15}+k\frac{\pi}{2}\ (TM)\ (k \in \mathbb{Z})`

Vậy `S={-\frac{4\pi}{15}+k\frac{\pi}{2}\ (k \in \mathbb{Z})}`

d) `cot\ 12x=1`

ĐK: `sin\ 12x \ne 0`

`⇔ 12x \ne k\pi\ (k \in \mathbb{Z})`

`⇔ x \ne \k\frac{\pi}{12}\ (k \in \mathbb{Z})`

`⇒ cot\ 12x=cot (\frac{\pi}{4})`

`⇔ 12x=\frac{\pi}{4}+k\pi\ (k \in \mathbb{Z})`

`⇔ x=\frac{\pi}{48}+k\frac{\pi}{4}\ (TM)\ (k \in \mathbb{Z})`

Vậy `S={\frac{\pi}{48}+k\frac{\pi}{4}\ (TM)\ (k \in \mathbb{Z})}`

e) `cot\ 5x =-\frac{1}{\sqrt{3}`

ĐK: `sin\ 5x \ne 0`

`⇔ 5x \ne k\pi\ (k \in \mathbb{Z})`

`⇔ x \ne \k\frac{\pi}{5}\ (k \in \mathbb{Z})`

`⇒ cot\ 5x=cot (\frac{2\pi}{3})`

`⇔ 5x=\frac{2\pi}{3}+k\pi\ (k \in \mathbb{Z})`

`⇔ x=\frac{2\pi}{15}+k\frac{\pi}{5}\ (TM)\ (k \in \mathbb{Z})`

Vậy `S={\frac{2\pi}{15}+k\frac{\pi}{5}\ (TM)\ (k \in \mathbb{Z})}`

f) `cot\ (3x+10^{0})=\frac{\sqrt{3}}{3}`

ĐK: `sin\ (3x+10^{0}) \ne 0`

`⇔ 3x+10^{0} \ne k180^{0}\ (k \in \mathbb{Z})`

`⇔ x \ne k\frac{170^{0}}{3}\ (k \in \mathbb{Z})`

`⇒ cot\ (3x+10^{0})=cot\ 60^{0}`

`⇔ 3x+10^{0}=60^{0}+k180^{0}\ (k \in \mathbb{Z})`

`⇔ x=\frac{50^{0}}{3}+k60^{0}\ (TM)\ (k \in \mathbb{Z})`

Vậy `S={\frac{50^{0}}{3}+k60^{0}\ (TM)\ (k \in \mathbb{Z})}`