Giải phương trình : cos6x * cos2x =sin7* sin3x

Đáp án:  $$\left[\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{9}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.$$

Giải thích các bước giải:

$\cos6x.\cos2x=\sin7x.\sin3x$

$\Leftrightarrow \dfrac{1}{2}\Big( \cos8x+\cos4x\Big)=\dfrac{-1}{2}\Big(\cos10x-\cos4x\Big)$

$\Leftrightarrow \cos8x+\cos4x=-\cos10x+\cos4x$

$\Leftrightarrow \cos10x+\cos8x=0$

$\Leftrightarrow 2\cos9x.\cos x=0$

$\Leftrightarrow  $ $$\left[\begin{matrix}\cos9x=0\\\cos x=0\end{matrix}\right.$$

$\Leftrightarrow$ $$\left[\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{9}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.$$

Đáp án:

\(
\left[ {\begin{array}{*{20}c}
   {x = \frac{\pi }{{18}} + \frac{{k\pi }}{9}}  \\
   {x = \frac{\pi }{2} - l\pi }  \\
\end{array}} \right.
\)
 Giải thích các bước giải:

Theo bài ra ta có: 

\(
\begin{array}{l}
 \cos 6x.\cos 2x = \sin 7x.\sin 3x \\ 
  \Leftrightarrow \frac{1}{2}\left( {\cos (6x + 2x) + \cos (6x - 2x)} \right) = \frac{1}{2}(\cos (7x - 3x) - c{\rm{os}}(7x + 3x)) \\ 
  \Leftrightarrow c{\rm{os}}8x + \cos 4x = c{\rm{os}}4x - c{\rm{os}}10x \\ 
  \Leftrightarrow \cos 8x =  - \cos 10x = \cos (\pi  - 10x) \\ 
  \Leftrightarrow \left[ {\begin{array}{*{20}c}
   {8x = \pi  - 10x + k2\pi }  \\
   {8x = 10x - \pi  + l2\pi }  \\
\end{array}} \right. \\ 
  \Leftrightarrow \left[ {\begin{array}{*{20}c}
   {18x = \pi  + k2\pi }  \\
   {2x = \pi  - l2\pi }  \\
\end{array}} \right. \\ 
  \Leftrightarrow \left[ {\begin{array}{*{20}c}
   {x = \frac{\pi }{{18}} + \frac{{k\pi }}{9}}  \\
   {x = \frac{\pi }{2} - l\pi }  \\
\end{array}} \right. \\ 
 \end{array}
\)