cos x +cos 2x +cos 3x =1/4?

Question

in this question the value of x has to be found out. solve.???

Answer 1

The answer is between 42.1 and 42.2 degrees, but I haven't found out how to get at it.

[edit]
This is as far as I can get right now.

using the relations:
cos 2x = 2cos^2 x - 1
cos 3x = 4cos^3 x - 3cos x

you get:
cos x + 2cos^2 x - 1 + 4cos^3 x - 3 cos x = 1/4
=> 2cos^3 x + cos^2 x - cos x = 5/8

let y = cos x

then,
2y^3 + y^2 - y - 5/8 = 0

factor and solve for y.
I know this equation is correct because if you put 42.11 in for x, ,then y = .7419 which works.

Once you find y, then x = arccos(y)

That's all I have time for right now...

Answer 2

I used a spreadsheet to find this out. One possible answer is:

x = 0.7349842943 radians, which equals 42.1114980718 degrees.

Plugging these into the equation gives 0.2500000000.

To get more precise than 0.2500000000 you'd need to use even more decimal points for x.

since the cos function is periodic, there is an infinite number of solutions to this equation

Answer 3

cosx+cos3x=2 cos 2xcosx and cosx+cos2x+cos3x
=cos2x(2cosx+1)=(2cos^2x-1)(2cosx+1)=4cos^3x+2cos^2x-2cosx-1=1/4 =>4cos^3x+2cos^2x-2cosx=5/4
=>2cos^3x+cos^2x-cosx=5/8 solving this for cosx
cosx=0.743 whih gives x as 42.1 deg.

Answer 4

cos 2x = 2cos^2x-1

cos3x = 4cos^3x-3cosx

Let cos x = y

So, 4y^3-3y+2y^2-1+y = 1/4
or 4y^3+2y^2+2y = 5/4
or 2y(2y^2+y+1)=5/4
The roots are complex.

Answer 5

you have to use your triginomic identities, which i dont remember off the top of my head. They should be in the back of your book

Answer 6

x=0.735 radian