Re: trig equations

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Posted by Joel on September 12, 2002 at 18:00:42:

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
**************************************************
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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Sorry, I'll try that again... Joel 18:04:21 09/12/02 (0)

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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 + 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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