solving trig equations
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sin^2x-cos^2x=0
1-cos^2x-cos^2x=0 1-2cos^2x=0 -2cos^2x=-1 cos^2x=1/2 sqaure route of cos^2x=square route of 1/2 cosx= + and- square route of 2/2 which equals 45, 135,225, and 315
4cos^2x-1=0
4cos^2x-1=0 4cos^2x=1 cos^2x=1/4 square route of cos ^2x= squARE route of 1/4 cos x= + or - 1/2
dont forget that if you have cos = 1/2 and cos = -1
they will each have different angle measures
when you ahve 3 terms to factor
use algera way of factoring and then put in the cos x
sin 2x =4sin x
use double Id 2sinxcosx-4sinx=0 2sinx (cosx-2)=0 2sinx=0 and cosx-2=0 sinx=0 and cosx=2 sinx=0 is 0 180 and 360 cosx2 is nothing
