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number of real solution of $\sin x\cdot \sin 2x \cdot \sin 3x = 1$ for all $x\in $ set of real numbers

$\sin x\cdot 2\sin x\cos x \cdot (3\sin x-4\sin^3 x) = 1$

$2\sin^3 x\cos x(3-4\sin^2 x) = 1$

$2\sin^3 x(3-4\sin^2 x) = \sec x$

drawing graph of LHS is hard, i want solution without graph

could some help me with this, thanks

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    I don't think this has real solutions as $\;|\sin x|\le 1\;$ for real $\;x\;$ .... – 2017-01-01
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    As a refinement, prove that $\forall x\in\mathbb{R} \ |\sin(x) \sin(2x)| \le \sqrt{16/27}$. – 2017-01-01
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    @ Gribouillis you mean $\displaystyle |\sin x\cdot \sin 2x \cdot \sin 3x| \leq |\sin x\cdot \sin 2x|\leq \frac{4}{3\sqrt{3}}$ – 2017-01-01
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    Yes it is a consequence, I used $16/27$ because one clearly sees that it is $< 1$ ! – 2017-01-01

3 Answers 3

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Since $|\sin x|\leq 1$, this is equivalent to $\sin x=\pm 1$ and $\sin (2x)=\pm 1$ and $\sin(3x)=\pm 1$ with zero or two $-1$. For zero $-1$'s, we have $$x=\pi/2+2\pi n=\pi/4+\pi m,$$ which is a contradiction. Similar for two $-1$'s. There are no solutions.

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Hint: Since $|\sin x| \le 1$, this can only be true if all of the factors have magnitude $1$ and $1$ or $3$ of them are positive.

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HINT:

For real $y,$

$$|\sin y|\le1$$

So, we need even number terms among $\sin x,\sin2x,\sin3x$ with value $=-1$

What is $\sin2x=2\sin x\cos x$ if $\sin x=\pm1?$