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? 
 A: Since
$$ \sin a \cdot \sin b = \frac{1}{2} \left[ \cos (a - b) - \cos (a+b) \right] $$
$$ \sin a \cdot \cos b = \frac{1}{2} \left[ \sin (a + b) + \sin (a - b) \right] $$
and
$$ \sin a - \sin b = 2 \sin\left[ \frac{1}{2} (a-b) \right] \cos \left[ \frac{1}{2} (a+b) \right] $$
we can use these identities to solve
$$ 4 \sin x \cdot \sin 2x \cdot \sin 4x = \sin 3 x $$
$$ \Rightarrow \quad 2 \sin x \cdot \left[ \cos (-2x) - \cos 6x \right] = \sin 3x $$
$$ \Rightarrow \quad 2 \sin x \cdot \cos 2x  - 2 \sin x \cdot \cos 6x = \sin 3x $$
$$ \Rightarrow \quad \sin 3x + \sin (-x)  - (\sin 7x + \sin(-5x)) = \sin 3x $$
$$ \Rightarrow \quad \sin 3x - \sin x  - \sin 7x + \sin 5x = \sin 3x $$
$$ \Rightarrow \quad - \sin x  - \sin 7x + \sin 5x = 0 $$
$$ \Rightarrow \quad \sin x = \sin 5x - \sin 7x $$
$$ \Rightarrow \quad \sin x = 2 \sin (-x) \cos 6x $$
$$ \Rightarrow \quad \sin x = - 2 \sin x \cos 6x $$
$$ \Rightarrow \quad \cos 6x = - \frac{1}{2} $$
$$ \Rightarrow \quad 6x = \frac{2}{3} (3 \pi n \pm \pi) $$
$$ \Rightarrow \quad x = \frac{1}{9} (3 \pi n \pm \pi) $$
or
$$\sin x = 0 \quad \Rightarrow \quad x = n \pi \mathrm{.}$$
A: The following statements are equivalent:


*

*$4\sin x\sin2x\sin4x=\sin3x$

*$4\left[\frac{1}{2i}\left(e^{ix}-e^{-ix}\right)\right]\left[\frac{1}{2i}\left(e^{2ix}-e^{-2ix}\right)\right]\left[\frac{1}{2i}\left(e^{4ix}-e^{-4ix}\right)\right]=\frac{1}{2i}\left(e^{3ix}-e^{-3ix}\right)$

*$-\left(e^{ix}-e^{-ix}\right)\left(e^{6ix}-e^{-2ix}-e^{2ix}+e^{-6ix}\right)=\left(e^{ix}-e^{-ix}\right)\left(e^{2ix}+1+e^{-2ix}\right)$

*$\left(e^{ix}-e^{-ix}\right)\left[e^{6ix}+e^{-6ix}+1\right]=0$

*$\sin x\left(2\cos6x+1\right)=0$
Solving the last equality is a lot easyer than solving the first.
The formula of the Moivre can be very helpful for questions like this.
A: \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:
$$x \in (\{60^\circ n  \pm 40^\circ : n \in \mathbb Z \} 
         \cup \{180^\circ n : n \in \mathbb Z\})
         \cap [0^\circ, 360^\circ]$$
$$x \in
   \left\{ \begin{array}{rrrrr}
        0^\circ, &  20^\circ, &  40^\circ, &  80^\circ, & 100^\circ, \\
      140^\circ, & 160^\circ, & 180^\circ, & 200^\circ, & 220^\circ, \\
      260^\circ, & 280^\circ, & 320^\circ, & 340^\circ, & 360^\circ \\
   \end{array} \right\} $$
A: Using  $\sin3B=3\sin B-4\sin^3B$
$$32\sin^3x\cdot\cos^2x\cdot\cos2x=\sin x(3-4\sin^2x)$$
If $\sin x=0,x=180^\circ n$ where $n$ is any integer
Else using $\cos2A=2\cos^2A-1=1-2\sin^2A,$
$$32\cdot\dfrac{1-\cos2x}2\cdot\dfrac{1+\cos2x}2\cdot\cos2x=3-2(1-\cos2x)$$
$$\iff8(c-c^3)=1+2c\iff4c^3-3c=-\dfrac12$$
$$\implies\cos6x=-\dfrac12,6x=360^\circ m\pm120^\circ\iff x=60^\circ m\pm20^\circ$$ where $m$ is any integer
