Sorry, I'll try that again...


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Posted by Joel on September 12, 2002 at 18:04:21:

In Reply to: Re: trig equations posted by Joel on September 12, 2002 at 18:00:42:

: : Thanks for all your help in the past!!!
: : solve without calculator

: : cos2x= sqrt3 / 2
: *************************************************
: : 2cos^2(x) - 1 = sqrt3 / 2 ..........use parentheses = less confusion
: : Then do you add the 1 ................yes but you forgot the right side

: : 2cos^2(x) = sqrt3 + 1
: **************************************************
: Now you fix the rest:

:
: : 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




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