<s>[INST] Solve for $x$ where $0\leq x\leq 360$ Solve 
$$4\sin x \cdot \sin 2x \cdot \sin 4x =\sin 3x$$
My Attempt :
Here,
$$4\sin x \cdot  \sin 2x \cdot  \sin 4x=\sin 3x$$
$$4\sin x \cdot (2\sin x  \cdot \cos x ). (4 \sin x  \cdot  \cos x  \cdot \cos 2x)=\sin3x$$
$$32\sin^3 x \cdot \cos^2 x \cdot \cos2x=\sin3x$$
How should I proceed further? 
 [/INST] \begin{align}
   \sin 3x 
   &= \sin(2x + x) \\
   &= \sin 2x \; \cos x + \cos 2x \; \sin x \\
   &= 2 \sin x \; \cos^2 x + \cos 2x \; \sin x \\
   &= \sin x \; (2 \cos^2 x + \cos 2x) \\
   &= \sin x \; (2 \cos 2x + 1)
\end{align}
\begin{align}
   4 \sin 2x \; \sin 4x
   &= 2(\cos 4x \; \cos 2x + \sin 4x \; \sin 2x)
     -2(\cos 4x \; \cos 2x - \sin 4x \; \sin 2x) \\
   &= 2\cos(4x - 2x) - 2\cos(4x + 2x) \\
   &= 2(\cos 2x - \cos 6x)
\end{align}
\begin{align}
   4\sin x \; \sin 2x \; \sin 4x &= \sin 3x \\
   2\sin x \; (\cos 2x - \cos 6x) &= \sin x \; (2 \cos 2x + 1) \\
\hline
   \sin x &= 0\\
   x &\in \{180^\circ n : n \in \mathbb Z\} \\
\hline
   2 \cos 2x - 2 \cos 6x &= 2 \cos 2x + 1 \\
   \cos 6x &= -\dfrac 12\\
   6x &\in \{360^\circ n \pm 240^\circ : n \in \mathbb Z \} \\
   x &\in \{60^\circ n  \pm 40^\circ : n \in \mathbb Z \} \\
\end{align}
Solution set:
$$x \in (\{60^\circ n  \pm 40^\circ : n \in \mathbb Z \} 
         \cup \{180^\circ n : n \in \mathbb Z\})
         \cap [0^\circ, 360^\circ]$$
$$x \in
   \left\{ \begin{array}{rrrrr}
        0^\circ, &  20^\circ, &  40^\circ, &  80^\circ, & 100^\circ, \\
      140^\circ, & 160^\circ, & 180^\circ, & 200^\circ, & 220^\circ, \\
      260^\circ, & 280^\circ, & 320^\circ, & 340^\circ, & 360^\circ \\
   \end{array} \right\} $$
</s>