Posted by Joel on September 12, 2002 at 20:20:41:
In Reply to: Re: trig equations posted by Subhotosh Khan on September 12, 2002 at 18:49:13:
: : Thanks for all your help in the past!!!
: : solve without calculator
: : cos2x= sqrt3 / 2
: : 2cos^2x-1 = sqrt3 / 2
: : Than do you add the 1
: : 2cos^2x = sqrt3
: : divide the 2
: ****************************
: Cos(2x) = sqrt(3)/2 = cos (pi/6+ 2*n*pi)
: where n = 0,1,2,3 etc.
: 2x = pi/6+ 2*n*pi
: x = pi/12 + n*pi
: : cos^2x= sqrt3 / 2
: : sqaure root both sides
: : cosx = 3 sqrt2 / 2
: : if this is right- which I believe it is not- I still did not come up with the right answers
: : Any help I will be thinkful!!!!!! KIM
Guess I was too worried about parentheses & arithmetic to think about what I was doing,as opposed to Mr. K, who comes up with a useful answer! :)