Asked by Sarah
Solve the equation cos3x=cos2x for 0¡Üx¡Ü360
Answers
Answered by
Steve
cos3x = cos2x
we need 3(x+2m*pi) = 2(x+2n*pi) for some m and n. Some playing around will get you
x = 0,2/5 pi, 4/5 pi
or, in degrees,
x = 0,144,288
we need 3(x+2m*pi) = 2(x+2n*pi) for some m and n. Some playing around will get you
x = 0,2/5 pi, 4/5 pi
or, in degrees,
x = 0,144,288
Answered by
Sarah
?? but the ans is 0, 72 ,144,216,288,360 degree
Answered by
Reiny
In terms of cosx
cos 3x = 4 cos^3 x - 3cosx , and cos 2x = 2 cos^2 x - 1
so cos 3x = cos 2x
4 cos^3 x - 3cosx - 2cos^2 x + 1 = 0
let cosx = y
so we need
4y^3 - 2y^2 - 3y + 1 = 0
I tried y = 1 and sure enough that was a solution.
then by synthetic division ...
4y^3 - 2y^2 - 3y + 1 = 0
(y-1)((4y^2 + 2y - 1) = 0
y = 1 or y = .30902 or y = -.80902 , using the formula
if y = 1, cosx = 1
x = 0 , 360
if y = .30902 , cosx = .30902
x = 72°, 288
if y = -.80902 , cosx = -.80902
x = 144° , 216°
x = 0, 72, 144, 216, 288, and 360°
cos 3x = 4 cos^3 x - 3cosx , and cos 2x = 2 cos^2 x - 1
so cos 3x = cos 2x
4 cos^3 x - 3cosx - 2cos^2 x + 1 = 0
let cosx = y
so we need
4y^3 - 2y^2 - 3y + 1 = 0
I tried y = 1 and sure enough that was a solution.
then by synthetic division ...
4y^3 - 2y^2 - 3y + 1 = 0
(y-1)((4y^2 + 2y - 1) = 0
y = 1 or y = .30902 or y = -.80902 , using the formula
if y = 1, cosx = 1
x = 0 , 360
if y = .30902 , cosx = .30902
x = 72°, 288
if y = -.80902 , cosx = -.80902
x = 144° , 216°
x = 0, 72, 144, 216, 288, and 360°
Answered by
Steve
Good catch, Reiny - I was a bit sloppy.
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