Posted by Subhotosh Khan on September 12, 2002 at 18:49:13:
In Reply to: trig equations posted by CAVANAUGH on September 12, 2002 at 16:11:08:
: 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