Câu 5:

$\begin{array}{l} \sqrt 3 \left( {\cos 7x - \cos 9x} \right) = \sin 9x - \sin 7x\\  \Leftrightarrow \sin 7x + \sqrt 3 \cos 7x = \sin 9x + \sqrt 3 \sin 9x\\  \Leftrightarrow \sin \left( {7x + \dfrac{\pi }{3}} \right) = \sin \left( {9x + \dfrac{\pi }{3}} \right)\\  \Leftrightarrow \left[ \begin{array}{l} 9x + \dfrac{\pi }{3} = 7x + \dfrac{\pi }{3} + k2\pi \\ 9x + \dfrac{\pi }{3} = \pi  - 7x - \dfrac{\pi }{3} + k2\pi  \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ x = \dfrac{\pi }{{48}} + \dfrac{{k\pi }}{8} \end{array} \right.\left( {k \in Z} \right)\\ k = 0 \Rightarrow x = 0(L),x = \dfrac{\pi }{{48}} \in \left( {0;\dfrac{\pi }{{12}}} \right)\\  \to \text{Nghiệm dương nhỏ nhất}\,x = \dfrac{\pi }{{48}} \in \left( {0;\dfrac{\pi }{{12}}} \right) \end{array}$

Câu 6:

$\begin{array}{l}
\sin x + \sqrt 3 \cos x = \sqrt 2 \\
 \Leftrightarrow 2\sin \left( {x + \dfrac{\pi }{3}} \right) = \sqrt 2 \\
 \Leftrightarrow \sin \left( {x + \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = \dfrac{\pi }{4} + k2\pi \\
x + \dfrac{\pi }{3} = \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{{12}} + k2\pi \\
x = \dfrac{{5\pi }}{{12}} + k2\pi 
\end{array} \right.\left( {k \in \mathbb Z} \right)\\
 \Rightarrow \alpha  = \dfrac{{ - \pi }}{{12}},\beta  = \dfrac{{5\pi }}{{12}}\\
 \Rightarrow \alpha \beta  =  - \dfrac{{5{\pi ^2}}}{{144}}
\end{array}$

#LC

Câu 5:

`\sqrt3 (cos7x-cos9x) = sin9x - sin7x`

⇔ `\sqrt3 cos7x + sin7x = \sqrt3 cos9x + sin9x`

⇔ `(\sqrt3)/(2) cos7x + 1/2 sin7x = (\sqrt3)/(2) cos9x + 1/2sin9x`

⇔ ` cos(π/6).cos7x + sin(π/6).sin7x = cos(π/6).cos9x + sin(π/6).sin9x`

⇔ `cos(π/6-7x) = cos(π/6-9x)`

⇔ \(\left[ \begin{array}{l}\frac{π}{6}-7x=\fracπ6-9x+k2π\\\frac{π}{6}-7x=-\fracπ6+9x+k2π\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=kπ\\x=\fracπ{48}-k\fracπ8\end{array} \right.\)

⇒ `π/(48)` là nghiệm dương nn: `π/(48) ∈ (0;π/(12))` 

⇒ Chọn A

Câu 6:

`sinx + \sqrt3 cosx= \sqrt2`

⇔ `1/2 sinx + (\sqrt3)/2 cosx = (\sqrt2)/2`

⇔ `cos(π/3).sinx + sin(π/3).cosx = sin(π/4)`

⇔ `sin(x+π/3) = sin(π/4)`

⇔ \(\left[ \begin{array}{l}x+\fracπ3=\fracπ4+k2π\\x+\fracπ3=\frac{3π}4+k2π\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{-π}{12}+k2π\\x=\frac{5π}{12}+k2π\end{array} \right.\) 

⇒ `\alpha=-π/12` và `\beta=(5π)/12`

⇒ `\alpha . \beta = (-π^2)/(144)

⇒ Chọn D