Trig help?

Question

cos^2 - 2 cosx - 1 = 0

I know the trick is to factor...but I cant seem to factor it properly.

Help, please?

Answer 1

Is it possible that the question should be:-
2 cos ² x - cos x - 1 = 0
(2 cos x + 1)(cos x - 1) = 0
cos x = - 1/2 , cos x = 1
x = 120° , 240° , 0° , 360°

Answer 2

Guessing that you accidentally left out the x in the cos^2. To factor this problem, start with cosx^2. You know that will be cosx * cosx. So the first factor in each () is cosx. Then look at the last factor -- it is a -1. To get a negative you have to multiply + * -. So the second factor is +1 in one () and -1 in the other (). So the result looks like this:
(cosx+1)(cosx-1) Then you must check your work. To do that, use the acronym FOIL (first, outer, inner, last). Multiplying the first give cosx^2, multiplying outer gives +1cosx, multiplying inner gives -1cosx [Houston we have a problem], and multiplying the last factor gives -1. The result would be cosx^2 -1. That's not right! How can we fix it? We can't change the -1 or the +1 or that will change the end result. So the only thing left to change is the first factor. If we change one to - and leave the other + then the result will be -cosx^2. But if we change both of the first factors to a - then - * - = + so the result is:
(-cosx^2 + 1)(-cosx2-1)
checking our work using the FOIL formula we see that the answer is correct!

Good Luck

Answer 3

parametrically let t = cos x then u have t^2-2t -1=0
by quadratic formula t = 1+root 2 or 1- root 2
cosx = 1+root 2 which is out of the domain of cos
or cox = 1-root 2 which mean x= 2 radians apprx
r u sure problem wasnt
cos^2(x) - 2cos(x) +1 =0 then cosx = 1 or x =0 radians

Answer 4

assume you mean
[cos(x)]^2 - 2cos(x) - 1 = 0

let z = cos(x) then we can rewrite this as
z^2 - 2z - 1 = 0
there is no way to factor this ... but you can solve for z using the quadratic forumla ...
az^2 + bz + c = 0
z = (-b + sqrt(b^2 - 4ac))/2a or
z = (-b - sqrt(b^2 - 4ac))/2a
in our case a=1, b=-2 and c=-1 so we have
z = 1 + sqrt(2) or 1 - sqrt(2)

now to solve for x in the original equation ...
since we know -1 <= cos(x) <= +1 we know that
z = 1+sqrt(2) CANNOT be a solution, therefore we are left with z = 1 - sqrt(2) = 1 - 1.414 = -0.414

x = arccos(-0.414) = 114.5 deg = 1.998 rad
(there are other answers because cos is a period function, but these are the smallest values of x for which solve the original equation)

Answer 5

cos^2 - 2 cosx = 1
cosx(cosx -2) = 1
go from there

Answer 6

cos^2 - 2 cosx - 1 = 0



let's add +2 to both sides of the equation to get
2+cos^2(x) - 2 cosx - 1 = 2+0

cos^2(x) -2cos(x) +1 = 2
cos^2(x) -2cos(x) +1 = (cos(x) -1)^2

hence,
(cos(x) -1)^2 =2

(cos(x) -1)^2 -2 =0
(cos(x) -1)^2 - [sqrt(2)]^2 = 0


this is in the form of (a^2 -b^2) =(a-b)(a+b)

{cos(x)-1 -sqrt(2)}{cos(x)-1 +sqrt(2)} =0

then you solve for x
either
cos(x)-1 -sqrt(2) =0 ---> cos(x) = 1+sqrt(2)
or
cos(x)-1 +sqrt(2) =0 ---> cos(x) = 1- sqrt(2)

Answer 7

Use quadratic formula. But use cos x on left side instead of x.

cos x = quadratic formula

Then when the formula is simplified, take the cosInv of the answer to get the angles.

Answer 8

(cos (x) -1) ^2 = 0

cos(x) = 1

x = 0 degrees in one full circle or x = 360 n for integer values on n.

In radians, x = 0 rad or 2n Pi for multiple rotations.