Trig equation

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Promilla

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Feb 22, 2012
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I was trying to solve this, but calculations are horrible :
sin2x*sin4x*sin6x=1
I have something like :
2sinxcosx*4sinxcosx(1-2sin^{2}x)*[2sinxcosx*cos4x+(1-2sin^{2}x)sin4x]=1
I think there must be simpler way to solve this .
Thanks in advance .
 
Promilla said:
I was trying to solve this, but calculations are horrible :
sin2x*sin4x*sin6x=1
I have something like :
2sinxcosx*4sinxcosx(1-2sin^{2}x)*[2sinxcosx*cos4x+(1-2sin^{2}x)sin4x]=1
I think there must be simpler way to solve this .
Thanks in advance .
Click to expand...
This is not easy. But break it down in bits.
\(\displaystyle \sin(4x)=2\sin(2x)\cos(2x)\)
\(\displaystyle \sin(6x)=\sin(2x+4x)=\sin(2x)\cos(4x)+\sin(4x)\cos(2x)\).
Can you carry on?
 
What did you decide about the DOMAIN before you started playing with the algebra? What do you know about the RANGE?

\(\displaystyle -1 \le sin(n*x) \le 1\)

This should suggest to you something fundamental about this problem. The ONLY way to get one (1) from it is one of these:

(1)(1)(1)
(1)(-1)(-1)
(-1)(1)(-1)
(-1)(-1)(1)

A more interesting task than all the algebra might be determining if any of these conditions ever occurs.
 
Ok . Thank you for the hint . So this means that I have to examine three cases and find out for what value of x it is possible right ?
I will try to do something with this.
 
?? What did you do that firghtened you away?

Just looking on \(\displaystyle [0,2\pi]\)

y = sin(x) takes on the value y = 1 at \(\displaystyle x = \frac{\pi}{2} + 2k\pi \) where k is an integer. Only k = 0 is in the desired Domain.

And y = -1 at \(\displaystyle x = \frac{3\pi}{2} + 2k\pi\) where k is an integer. Again, only k = 0 is in the desired Domain

Now you do y = sin(2x). Where does it take on y = 1 and where does it take on y = -1?
 
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