Đáp án:

a. \(\left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x =  - \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{3} + k\pi \\
x =  - \dfrac{\pi }{3} + k\pi 
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\left[ \begin{array}{l}
\cos 2x = \dfrac{1}{2}\\
\cos 2x =  - \dfrac{1}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x = \dfrac{\pi }{3} + k2\pi \\
2x =  - \dfrac{\pi }{3} + k2\pi \\
2x = \dfrac{{2\pi }}{3} + k2\pi \\
2x =  - \dfrac{{2\pi }}{3} + k2\pi 
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x =  - \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{3} + k\pi \\
x =  - \dfrac{\pi }{3} + k\pi 
\end{array} \right.\left( {k \in Z} \right)\\
b.DK:\cos x \ne 0 \to x \ne \dfrac{\pi }{2} + k\pi \\
\left[ \begin{array}{l}
\cos 2x = 0\\
\tan x = 0
\end{array} \right. \to \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
x = k\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = k\pi 
\end{array} \right.\\
 \to x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
c.DK:\sin x \ne 0 \to x \ne k\pi \\
\sin 3x.\cot x = 0\\
 \to \left[ \begin{array}{l}
\sin 3x = 0\\
\cos x = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
3x = k\pi \\
x = \dfrac{\pi }{2} + k\pi 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
x = \dfrac{\pi }{2} + k\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)

Đáp án:

`a)` \(\left[ \begin{array}{l}x = ±\frac{π}{6}\\x = ±\frac{π}{3}\end{array} \right.\) `(k ∈ ZZ)`

`b)` \(\left[ \begin{array}{l}x = \frac{π}{4} + k\frac{π}{2}\\x = kπ\end{array} \right.\) `(k ∈ ZZ)`

`c)`  \(\left[ \begin{array}{l}x = k\frac{π}{3}\\x = \frac{π}{2} + kπ\end{array} \right.\) `(k ∈ ZZ)`

Giải thích các bước giải:

`a) cos² 2x = 1/4`

`<=>`  \(\left[ \begin{array}{l}cos 2x = frac{1}{2}\\cos 2x = -\frac{1}{2}\end{array} \right.\) 

`<=>`  \(\left[ \begin{array}{l}2x = ±\frac{π}{3}\\2x = ±\frac{2π}{3}\end{array} \right.\) 

`<=>`  \(\left[ \begin{array}{l}x = ±\frac{π}{6}\\x = ±\frac{π}{3}\end{array} \right.\) `(k ∈ ZZ)`

`b) cos 2x.tan x = 0`

`ĐKXĐ: x ne (π)/2 + kπ`

`=>`  \(\left[ \begin{array}{l}cos 2x = 0\\tan x = 0\end{array} \right.\) 

`<=>`  \(\left[ \begin{array}{l}2x = \frac{π}{2} + kπ\\sin x = 0\end{array} \right.\) 

`<=>`  \(\left[ \begin{array}{l}x = \frac{π}{4} + k\frac{π}{2}\\x = kπ\end{array} \right.\) `(k ∈ ZZ)`

`c) sin 3x.cot x = 0`

`ĐKXĐ: x ne kπ`

`=>` \(\left[ \begin{array}{l}sin 3x = 0\\cot x = 0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}3x = kπ\\x = \frac{π}{2} + kπ\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x = k\frac{π}{3}\\x = \frac{π}{2} + kπ\end{array} \right.\) `(k ∈ ZZ)`