Đáp án:

$\begin{array}{l}6)\,\,x = \dfrac{\pi}{4} + \dfrac{1}{2}\arccos\dfrac{5}{13} + k\pi \quad (k \in \Bbb Z)\\7)\,\,\left[\begin{array}{l}x = \dfrac{\pi}{12} + k\pi\\x = \dfrac{\pi}{24}+ k\dfrac{\pi}{12}\end{array}\right.\quad (k \in \Bbb Z)\\8)\,\,x = \dfrac{1}{2}\arccos\dfrac{2}{\sqrt5} + k\dfrac{\pi}{2}\quad (k \in \Bbb Z)\end{array}$

Giải thích các bước giải:

$\begin{array}{l}6)\,\,5\sin2x - 12\cos2x = 13\\ \Leftrightarrow \dfrac{5}{13}\sin2x - \dfrac{12}{13}\cos2x = 1\\ Do\,\,\left(\dfrac{5}{13}\right)^2 + \left(\dfrac{12}{13}\right)^2 = 1\\ Đặt \,\,\begin{cases}\cos\alpha = \dfrac{5}{13}\\\sin\alpha = \dfrac{12}{13}\end{cases}\Rightarrow \alpha = \arccos\dfrac{5}{13}\\ \text{Phương trình trở thành:}\\ \sin2x.\cos\alpha - \cos2x,\sin\alpha = 1\\ \Leftrightarrow \sin(2x - \alpha) = 1\\ \Leftrightarrow 2x - \alpha = \dfrac{\pi}{2} + k2\pi\\ \Leftrightarrow 2x = \dfrac{\pi}{2} + \alpha + k2\pi\\ \Leftrightarrow x = \dfrac{\pi}{4} + \dfrac{1}{2}\alpha + k\pi\\ \Leftrightarrow x = \dfrac{\pi}{4} + \dfrac{1}{2}\arccos\dfrac{5}{13} + k\pi \quad (k \in \Bbb Z)\\ 7)\,\,\cos7x - \sin5x = \sqrt3(\cos5x - \sin7x)\\ \Leftrightarrow \dfrac{1}{2}\cos7x + \dfrac{\sqrt3}{2}\sin7x = \dfrac{3}{2}\cos5x + \dfrac{1}{2}\sin5x\\ \Leftrightarrow \cos\left(7x - \dfrac{\pi}{3}\right) = \cos\left(5x - \dfrac{\pi}{6}\right)\\ \Leftrightarrow \left[\begin{array}{l}7x - \dfrac{\pi}{3} = 5x - \dfrac{\pi}{6} + k2\pi\\7x - \dfrac{\pi}{3} = \dfrac{\pi}{6} - 5x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{6} + k2\pi\\12x = \dfrac{\pi}{2}+ k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k\pi\\x = \dfrac{\pi}{24}+ k\dfrac{\pi}{12}\end{array}\right.\quad (k \in \Bbb Z)\\ 8)\,\,\sin2x + \sin^2x = \dfrac{1}{2}\\ \Leftrightarrow \sin2x + \dfrac{1 - \cos2x}{2} = \dfrac{1}{2}\\ \Leftrightarrow 2\sin2x - \cos2x = 0\\ \Leftrightarrow \dfrac{2}{\sqrt5}\sin2x - \dfrac{1}{\sqrt5}\cos2x =0\\ Do\,\,\left(\dfrac{2}{\sqrt5}\right)^2 + \left(\dfrac{1}{\sqrt5}\right)^2 = 1\\ Đặt\,\begin{cases}\cos\alpha = \dfrac{2}{\sqrt5}\\\sin\alpha = \dfrac{1}{\sqrt5}\end{cases}\Rightarrow \alpha = \arccos\dfrac{2}{\sqrt5}\\ \text{Phương trình trở thành:}\\ \sin2x.\cos\alpha - \cos2x.\sin\alpha = 0\\ \Leftrightarrow \sin(2x - \alpha) = 0\\ \Leftrightarrow 2x - \alpha = k\pi\\ \Leftrightarrow 2x = \alpha + k\pi\\ \Leftrightarrow x = \dfrac{1}{2}\alpha + k\dfrac{\pi}{2}\\ \Leftrightarrow x = \dfrac{1}{2}\arccos\dfrac{2}{\sqrt5} + k\dfrac{\pi}{2}\quad (k \in \Bbb Z) \end{array}$

`6) 5cos 2x - 12cos 2x = 13`

`<=> -7cos 2x = 13`

`<=> cos 2x = -13/7`

`=>` Phương trình vô nghiệm

`7) cos 7x - sin 5x = sqrt{3}(cos 5x - sin 7x)`

`<=> cos 7x - sin 5x = sqrt{3}cos 5x - sqrt{3}sin 7x`

`<=> sqrt{3}sin 7x + cos 7x = sin 5x + sqrt{3}cos 5x`

`<=> sin (7x + π/6) = sin (5x + π/3)`

`<=>` \(\left[ \begin{array}{l}7x + \dfrac{π}{6} = 5x + \dfrac{π}{3} + k2π\\7x + \dfrac{π}{6} = π - 5x - \dfrac{π}{3} + k2π\end{array} \right.\) 

 `<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{12} + kπ\\x = \dfrac{π}{24} + k\dfrac{π}{6}\end{array} \right.\) `(k ∈ ZZ)`

`8) sin 2x + sin^2 x = 1/2`

`<=> sin 2x + (1 - cos 2x)/2 = 1/2`

`<=> 2sin 2x + 1 - cos 2x - 1 = 0`

`<=> 2sin 2x - cos 2x = 0`

Đặt: 

\(\left\{ \begin{array}{l}\dfrac{2}{\sqrt{5}} = cos ∝\\\dfrac{1}{\sqrt{5}} = sin ∝\end{array} \right.\) 

`=> cos ∝.sin 2x - sin ∝.cos 2x = 0`

`<=> sin (2x - ∝) = 0`

`<=> 2x - ∝ = kπ`

`<=> x = ∝/2 + k(π)/2` `(k ∈ ZZ)`