Here's what I got, but I'm still stuck:
2 sin x(1-sin^2x)=1-sin^2x-sin^2x
2 sin x - 2 sin^3 x=1 - 2 sin^2 x
2 sin^3 x - 2 sin^2 x - 2 sin x +1=0
I can't factor by grouping - I've tried every combination and it doesn't work.
I also thought I might be able to move the 1 to the other side and then factor, but that didn't work either.
2 sin x(sin^2 x - sin x - 1)=-1 but what is left inside the parentheses is not factorable.
Help!
The advice you've been given is good, and is standard. I'm wondering, though, if anybody else has tried following this through to the end. There are solutions (
here), but I'm having a devil of a time arriving at them independently.
The equation can be rearranged thusly:
. . . . .\(\displaystyle 2\, \sin(x)\, \cos^2(x)\, =\, \cos^2(x)\, -\, \sin^2(x)\)
. . . . .\(\displaystyle \sin^2(x)\, +\, 2\, \sin(x)\, \cos^2(x)\, -\, \cos^2(x)\, =\, 0\)
This is a quadratic in sine, so we can apply the Quadratic Formula:
. . . . .\(\displaystyle \sin(x)\, =\, \dfrac{-(2\, \cos^2(x))\, \pm\, \sqrt{(2\, \cos^2(x))^2\, -\, 4(1)(-\cos^2(x))\,}}{2(1)}\)
. . . . .\(\displaystyle \sin(x)\, =\, \dfrac{-2\, \cos^2(x)\, \pm\, \sqrt{4\, \cos^4(x)\, +\, 4\, \cos^2(x)\,}}{2}\)
...which eventually ends up in the vicinity of:
. . . . .\(\displaystyle \sin(x)\, =\, -\cos(x)\, \pm\, \cos(x)\,\sin(x)\)
This technique is sometimes helpful, but it doesn't appear to be, in this case. I'm open to corrections and further hints, but I'm grinding to a halt, too.
I don't think it's you (your skills or your imagination); this equation is in fact a bit of a bear.... :shock: