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Lukecrayonz
cos 4x - cos 2x = 0 Find all solutions to the equation in the interval [0, 2ð). ð, pi/3 2pi/3 4pi/3 5pi/3, 0? Is that correct?
\(\cos^4 x - \cos^2x = 0\) is the equation like above ?
No, Cos(4x)-cos(2x)=0
@ganeshie8 no idea?:(
could u show me ur work
cosx=cos(x+2kpi) Apply Cos C - Cos D formula Cos C - Cos D = 2sin(C+D)/2 Sin(D-C)/2
So 0, pi/3, 2pi/3, 4pi/3, 5pi/3 and ð
@anas2000 can you help?:D
hmm that doesnt help me
@UnkleRhaukus @robtobey
cos 4 x - cos 2 x = 0 2cos^2 2x-1 - cos 2x = 0 2u^2 -u -1 = 0 (u-1)(2u+1) = 0 u = 1; u = -1/2 cos 2x = 1 ; cos 2x = -1/2 2x = 0 ; 2x = pi-pi/3, 2x = pi + pi/3 x = 0 + npi; x = pi/3 + npi, x = 2pi/3 + npi x = 0, pi; x = pi/3, 4pi/3, x = 2pi/3, 5pi/3 so solutions will be : x = 0, pi/3, 2pi/3, 4pi/3, 5pi/3, pi looks you got the same solutions wid very less effort... !
np... yw :) good work !