Posted by cavanaugh on September 13, 2002 at 01:40:22:
In Reply to: Re: trig equations posted by Subhotosh Khan on September 12, 2002 at 18:49:13:
: : 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
So my first mistake is that I do not rewrite cos2x to 2cosx-1.
I forgot to say that it must be solve in the interval [0,2pi)
I understand how pi/12 is solved. But how did 11pi/12, 13pi/12, 23pi/12 come into the picture.
i know it was
180-15=165
180+15=195
360-15=345
I thought the cos is only positve- QuadI and Quad IV.
: : 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