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= 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

Re: trig equations Subhotosh Khan 18:49:13 09/12/02 (4)
- Re: trig equations cavanaugh 01:40:22 09/13/02 (2)
  - Re: trig equations Subhotosh Khan 08:25:33 09/13/02 (0)
  - Re: trig equations Subhotosh Khan 08:25:26 09/13/02 (0)
- Yes ... Joel 20:20:41 09/12/02 (0)
I wish we had erasers here. Ignore my 2 posts below :) Joel 18:07:43 09/12/02 (0)
Re: trig equations Joel 18:00:42 09/12/02 (1)
- Sorry, I'll try that again... Joel 18:04:21 09/12/02 (0)