I wish we had erasers here. Ignore my 2 posts below :)

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Posted by Joel on September 12, 2002 at 18:07:43:

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^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)/2 + 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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: : : : 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)/2 + 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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