kindly explain ..
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cos 2x +cos 4x +cos 6x = 0 .Find X.?
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assume 0° ≤ x ≤ 360°
cos 2x + cos 4x + cos 6x = 0
cos 4x + (cos 2x + cos 6x) = 0
cos 4x + (2 cos 4x cos 2x) = 0
(1 + 2 cos 2x) cos 4x = 0
(1 + 2 cos 2x) = 0
cos 2x = -½
2x = 120°
x = 60°
or
2x = 240°
x = 120°
or
cos 4x = 0
4x = 90°
x = 22.5°
or
4x = 270°
x = 67.5°
x = 22.5° , 60° , 67.5° , 120°
hi! i don't know the exact method of doing it but by reasoning and general analysis i got the answer.
let 2x=A
so the problem now is, cos A + cos(2A) + cos (3A)
if you see the cosine function, it is positive in the 1st and the 4th quadrants, i.e. from 0deg to 90 deg and from 270 deg to 360 deg. the cosine function is negative from 90 deg to 270 deg.
so an angle A is such that adding the cosine value of itself, its twice n thrice would amount to 0. so i assumed a value of A such that one of the three terms would be 1, which is the highest value of the cosine function. and the other two with value half with negative sign.
so A=120, cos A= -0.5
2A=240, cos 240=-0.5
3A=360, cos 3A=1.
therefore, -0.5-0.5+1=0
ANSWER- X=120 degrees
hope it helped
cos 2x +cos 4x +cos 6x = 0
cos 2x +cos4x+cos(2x+4x)=
cos 2x +cos 4x +cos2x.cos4x-sin2x.sin4x=0
cos 2x (cos4x+1)+cos4x-sin2x.sin4x=0
2cos^3 2x+cos4x-2sin^2 2x. cos2x=0
2cos 2x (cos^2 2x-sin^2 2x)+cos4x=0
2cos 2x cos 4X +cos4x=0
cos4x=0 or 2cos 2x +1 =0 (not possible)
so Cos4x=0=cos (2n+1) pi/2
x= (2n+1) pi/8