Đáp án:

$\begin{array}{l}4)\,\,\left[\begin{array}{l}x = \dfrac{\pi}{8} +k\dfrac{\pi}{2}\\x = \dfrac{1}{2}\arctan\dfrac{1}{3} + k\dfrac{\pi}{2}\end{array}\right.\quad (k \in \Bbb Z)\\5)\,\,\left[\begin{array}{l}x = \dfrac{\pi}{4} +k\pi\\x = \arctan\dfrac{1}{2}+ k\pi\end{array}\right.\quad (k\in \Bbb Z)\end{array}$

Giải thích các bước giải:

$\begin{array}{l}4) \,3\tan^22x - 4\tan x + 1 \qquad (*)\\ ĐKXĐ: \, \cos2x \ne 0 \Leftrightarrow x \ne \dfrac{\pi}{4} + n\dfrac{\pi}{2} \quad (n \in \Bbb Z)\\ (*) \Leftrightarrow (\tan2x - 1)\left(\tan2x - \dfrac{1}{3}\right) = 0\\ \Leftrightarrow \left[\begin{array}{l}\tan2x = 1\\\tan2x = \dfrac{1}{3}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{4} +k\pi\\2x = \arctan\dfrac{1}{3} + k\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{8} +k\dfrac{\pi}{2}\\x = \dfrac{1}{2}\arctan\dfrac{1}{3} + k\dfrac{\pi}{2}\end{array}\right.\quad (k \in \Bbb Z)\\ 5)\,2\sin^2x - 3\sin x\cos x + \cos^2x = 0\\ \text{Nhận thấy $\cos x = 0$ không là nghiệm của phương trình}\\ \text{Chia 2 vế của phương trình cho $\cos^2x$ ta được:}\\ 2\tan^2x - 3\tan x + 1 = 0\\ \Leftrightarrow (\tan x -1)\left(\tan x - \dfrac{1}{2}\right) = 0\\ \Leftrightarrow \left[\begin{array}{l}\tan x = 1\\\tan x = \dfrac{1}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} +k\pi\\x = \arctan\dfrac{1}{2}+ k\pi\end{array}\right.\quad (k\in \Bbb Z) \end{array}$

`3tan² 2x - 4tan 2x + 1 = 0`

`<=>` \(\left[ \begin{array}{l}tan 2x = 1\\tan 2x = \dfrac{1}{3}\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x = \dfrac{π}{4} + kπ\\2x = arctan \dfrac{1}{3} + kπ\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{8} + k\dfrac{π}{2}\\x = \dfrac{arctan \dfrac{1}{3}}{2}+ k\dfrac{π}{2}\end{array} \right.\) `(k ∈ ZZ)`

`2sin² x - 3sin x.cos x + cos²x = 0`

- Với `cos x = 0`

`=> 2 = 0` (không thoã mãn)

- Với `cos x ne 0`

`=> 2tan² x - 3tan x + 1 = 0`

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

`<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{4} + kπ\\x = arctan \dfrac{1}{2} + kπ\end{array} \right.\) `(k ∈ ZZ)`