Why doesn't the trigonometric Equation $\sin (x) \sin (2x) \sin(3x)=1$ have real solutions While practising some problems I encountered a question:
Prove that the following equation has no real solution,
$$\sin (x) \sin (2x) \sin(3x)=1$$
At first glance, I thought about applying Rolle's theorem or check the domain of the function but all these approaches lead me to a dead-end.
Could anyone provide an insight into the technique that could be used to handle such questions?
 A: If there is a solution then $|\sin(x)|=|\sin(2x)|=|\sin(3x)|=1$.
A: $|\sin(x)|\leq1$ for all $x$. Since $\sin x\sin 2x\sin 3x=1$, we need $|\sin x|=|\sin 2x|=|\sin 3x|=1$
Hence $x, 2x, 3x$ have to be of the form $\frac{(2k+1)\pi}{2}$, where $k$ is an integer.
Since $x$ is an odd multiple of $\frac{\pi}{2}$, $2x$ has to be an even multiple of $\frac{\pi}{2}$.
A: For real $x$, $-1\leq\sin(x)\leq 1$. To have a product of sines equal to one, all of them would have to be $1$, or two of them would have to be $-1$. In either case, one of them has to be $1$, which means one of the following is true for some integer $n$:
$$x=\pi/2+2\pi n$$
$$2x=\pi/2+2\pi n$$
$$3x=\pi/2+2\pi n.$$
If the first is true, then $2x=\pi+4\pi n$, and $\sin(2x)=0$, which is no good. If the second is true, then $x=\pi/4+n\pi$, and $|\sin(x)|\neq 1$. The same problem arises for if the third is true. Thus, there are no real solutions.
A: It must be the case that $\sin x = \pm 1$, so that $x$ is an odd multiple of $\frac{\pi}{2}$. But then $2x$ is a multiple of $\pi$, so that $\sin 2x = 0$.
A: As |sin⁡(x)|≤1 sin(x)sin(2x)sin(3x)=1 have a solution only in four cases: 


*

*sin(x)=1 && sin(2x)=1 && sin(3x)=1 

*sin(x)=1 && sin(2x)=-1 && sin(3x)=-1
or if sin(x)=1 =>  x= π/2 + 2kπ   ; k is some integer
   => 2x= π + 4kπ && 3x= 3π/2 +6kπ 
   =>sin(2x)=0 != 1   && sin(3x)=-1  !=1
   => impossible

*sin(x)=-1 && sin(2x)=1 && sin(3x)=-1 

*sin(x)=-1 && sin(2x)=-1 && sin(3x)=1 
or if sin(x)=-1 =>  x= 3π/2 + 2kπ   ; k is some integer
   => 2x= 3π + 4kπ && 3x= 9π/2 +6kπ
   =>sin(2x)=0 != 1   && sin(3x)=sin(9π/2)=sin(π/2)=1
   => impossible
So there is no solution for the equation sin(x)sin(2x)sin(3x)=1
Second method:
sin(x)sin(2x)sin(3x)=1
|sin(x)sin(2x)sin(3x)|=1
|sin(x)|=1 && |sin(2x)|=1 && |sin(3x)|=1
or |sin(x)|=1 => x=  π/2 + kπ   ; k is some integer
=> 2x=  π + 2kπ 
=>|sin(2x)|=sin(π)=0 != 1
=> impossible
