Posted by Subhotosh Khan on September 13, 2002 at 08:25:33:
In Reply to: Re: trig equations posted by cavanaugh on September 13, 2002 at 01:40:22:
: : : 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.
you need cos (2x) to be positive. So:
2x = 11(pi)/6 = 330 IVth quadrant etc.
Here you have two answers x = pi/12 and 11(pi)/12
: : : 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