To quote courtigrad:
" \sin 2x = 2\sin x \cos x "
I think this might be a problem for some, because I was doing problems similar to this the other day, and only today did we begin learning the double-angle, half-angle, and power down formulas.
Is it incorrect to work both sides of the problem and meet in the middle, or do you need to get one side to completely equal another, because I can easily get both sides to be 1/sinxcosx.
on 3 and 4 I am still having trouble, I can get 3 to:
cos3x=cosx
u3-u=0
cosx(cos2x-1)=0
cosx(sin2x)=0
then get stuck
and number 4 I can't figure past where you got me
Thanks a bunch, was a BIG help, I can already see myself loving this website, I have a mediocre teacher and haven't been doing the best on tests, even though I ace quizzes, thanks again!
1.) sin2x=3cos2
2.) (3tan2x-1)(tan2x-3)=0
3.) cos3x=cosx
4.) 3tan3x=tanx
Those were a few of the problems in the section I missed on friday due to illness and I was wondering if anyone could walk me through them. Would be greatly appreciated!
They want to solve the equation for...