Đáp án:

$x = \dfrac{\pi }{{15}} + k\dfrac{{2\pi }}{{15}}\left( {k \ne 15m + 7,m \in Z} \right)$

Giải thích các bước giải:

Ta có:

$\begin{array}{l}
\dfrac{1}{{\sin x}} = \dfrac{1}{{\sin 2x}} + \dfrac{1}{{\sin 4x}} + \dfrac{1}{{\sin 8x}}\left( {DK:\sin 8x \ne 0 \Leftrightarrow x \ne \dfrac{{k\pi }}{8}} \right)\\
 \Leftrightarrow \dfrac{1}{{\sin x}} + \cot 8x = \dfrac{1}{{\sin 2x}} + \dfrac{1}{{\sin 4x}} + \dfrac{1}{{\sin 8x}} + \cot 8x\\
 \Leftrightarrow \dfrac{{\cot 8x\sin x + 1}}{{\sin x}} = \dfrac{1}{{\sin 2x}} + \dfrac{1}{{\sin 4x}} + \dfrac{1}{{\sin 8x}} + \dfrac{{\cos 8x}}{{\sin 8x}}\\
 \Leftrightarrow \dfrac{{\cot 8x\sin x + 1}}{{\sin x}} = \dfrac{1}{{\sin 2x}} + \dfrac{1}{{\sin 4x}} + \dfrac{{1 + \cos 8x}}{{\sin 8x}}\\
 \Leftrightarrow \dfrac{{\cot 8x\sin x + 1}}{{\sin x}} = \dfrac{1}{{\sin 2x}} + \dfrac{1}{{\sin 4x}} + \dfrac{{2{{\cos }^2}4x}}{{2\sin 4x\cos 4x}}\\
 \Leftrightarrow \dfrac{{\cot 8x\sin x + 1}}{{\sin x}} = \dfrac{1}{{\sin 2x}} + \dfrac{1}{{\sin 4x}} + \dfrac{{\cos 4x}}{{\sin 4x}}\\
 \Leftrightarrow \dfrac{{\cot 8x\sin x + 1}}{{\sin x}} = \dfrac{1}{{\sin 2x}} + \dfrac{{1 + \cos 4x}}{{\sin 4x}}\\
 \Leftrightarrow \dfrac{{\cot 8x\sin x + 1}}{{\sin x}} = \dfrac{1}{{\sin 2x}} + \dfrac{{2{{\cos }^2}2x}}{{2\sin 2x\cos 2x}}\\
 \Leftrightarrow \dfrac{{\cot 8x\sin x + 1}}{{\sin x}} = \dfrac{1}{{\sin 2x}} + \dfrac{{\cos 2x}}{{\sin 2x}}\\
 \Leftrightarrow \dfrac{{\cot 8x\sin x + 1}}{{\sin x}} = \dfrac{{1 + \cos 2x}}{{\sin 2x}}\\
 \Leftrightarrow \dfrac{{\cot 8x\sin x + 1}}{{\sin x}} = \dfrac{{2{{\cos }^2}x}}{{2\sin x\cos x}}\\
 \Leftrightarrow \dfrac{{\cot 8x\sin x + 1}}{{\sin x}} = \dfrac{{\cos x}}{{\sin x}}\\
 \Leftrightarrow \cot 8x\sin x + 1 = \cos x\\
 \Leftrightarrow \dfrac{{\cos 8x}}{{\sin 8x}}.\sin x + 1 = \cos x\\
 \Leftrightarrow \cos 8x\sin x + \sin 8x = \cos x\sin 8x\\
 \Leftrightarrow \sin 8x = \cos x\sin 8x - \cos 8x\sin x\\
 \Leftrightarrow \sin 8x = \sin \left( {8x-x} \right)\\
 \Leftrightarrow \sin 8x = \sin 7x
\end{array}$ 

$\begin{array}{l}
 \Leftrightarrow \sin 8x = \sin 7x\\
 \Leftrightarrow \left[ \begin{array}{l}
8x = 7x + k2\pi \\
8x = \pi  - 7x + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \left( l \right)\\
x = \dfrac{\pi }{{15}} + k\dfrac{{2\pi }}{{15}}
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{\pi }{{15}} + k\dfrac{{2\pi }}{{15}}\left( {k \ne 15m + 7,m \in Z} \right)
\end{array}$

Vậy phương trình có họ nghiệm là:

$x = \dfrac{\pi }{{15}} + k\dfrac{{2\pi }}{{15}}\left( {k \ne 15m + 7,m \in Z} \right)$