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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?

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  • $\begingroup$ Multiply both sides by $\cos x$ to get: $$4\sin^2 2x \sin 4x=\sin 2x+\sin 4x$$ or $4a^2 b=a+b$ $\endgroup$
    – polfosol
    Oct 29 '16 at 8:20
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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 $$\sin x = 0 \quad \Rightarrow \quad x = n \pi \mathrm{.}$$

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  • $\begingroup$ Oh yeah, your $x$ is probably in degrees, remember to switch from radians ($\pi$ rad $= 180^\circ$). $\endgroup$
    – 655321
    Oct 29 '16 at 8:43
  • $\begingroup$ @StevenGregory Well, knowing that $\cos \frac{\pi}{3} = \cos \left( - \frac{\pi}{3} \right) = \frac{1}{2}$ one can deduce that $\cos \left( \pi \pm \frac{\pi}{3} \right) = \cos\left(\pm \frac{2\pi}{3} \right) = - \frac{1}{2}$ and more generally $\cos \left( \pm \frac{2 \pi}{3} \pm 2 \pi n \right) = \cos\left[ \frac{2}{3} \left( \pm \pi \pm 3 \pi n \right) \right] = - \frac{1}{2}$. Noting that $\cos z = \cos (-z)$ one can write $ \cos\left[ \frac{2}{3} \left( \pm \pi \pm 3 \pi n \right) \right] = \cos \left[\frac{2}{3} \left( 3 \pi n \pm \pi \right) \right] $. $\endgroup$
    – 655321
    Oct 30 '16 at 8:10
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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.

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\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:

$$x \in (\{60^\circ n \pm 40^\circ : n \in \mathbb Z \} \cup \{180^\circ n : n \in \mathbb Z\}) \cap [0^\circ, 360^\circ]$$

$$x \in \left\{ \begin{array}{rrrrr} 0^\circ, & 20^\circ, & 40^\circ, & 80^\circ, & 100^\circ, \\ 140^\circ, & 160^\circ, & 180^\circ, & 200^\circ, & 220^\circ, \\ 260^\circ, & 280^\circ, & 320^\circ, & 340^\circ, & 360^\circ \\ \end{array} \right\} $$

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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

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  • $\begingroup$ @Théophile, One set of solution is $$0\le180n\le360\iff0\le n\le2$$ $\endgroup$ Nov 2 '16 at 17:57

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