Question

find all solutions to the equation in the interval [0, 2pi).
cos4x-cos2x=0
options are
A) 0, 2pi/3,4pi/3
B) 0, pi/3, 2pi/3, pi, 4pi/3, 5pi/3.
C) pi/6, pi/2, 5pi/6, 7pi/6, 3pi/2, 11pi/6
D) no solution

Answer 1

hint:\[\cos(2x)=2\cos^2x-1\]

Answer 2

cos4x=2cos^2(2x)-1; note 2x = angle
thereore
2cos^2 (2x)-1-cos(2x)=0
cos(2x) [2cos(2x)-1]=0
if, cos(2x)=0 then cos^2 (x)-sin^2(x)=0
ONE OF THE SOLUTIONS is pi/4 or 45 degrees, find rest

else
cos(2x)= 1/2
2x= inverse cosine (1/2),
2x= pi/6, 11pi/6
therefore x= pi/12, 11 pi/12

Answer 3

اHell0
cos4x=cos2x
(4x=2x+2kpi ) or (4x=-2x+2k'pi)
x=kpi or x= 2kpi/3
k=o x=0
k=1 x=pi
k'=1 x= (2pi)/3
k'=2 x=(4pi)/3

Answer 4

Thanks everybody!