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This is a problem I cant seem to understand....

Find, to the nearest degreee, all values of θ (theta) in the interval:

0 < (or equal)θ<(or equal)180 that satisfy the equation

8 cos^2 θ-2cosθ-1=0

How would I go about doing this? Help is appreciated...

2007-12-27 08:21:45 · 6 answers · asked by Anonymous in Science & Mathematics Mathematics

6 answers

Hi,

60° or 104°

8cos² Ө - 2 cos Ө - 1 = 0
(4 cos Ө + 1)(2 cos Ө - 1) = 0
4 cos Ө + 1 = 0
4 cos Ө = -1
cos Ө = -¼
cos^-1(-¼) = 104°

2 cos Ө - 1 = 0
2 cos Ө = 1
cos Ө = ½
cos^-1(½) = 60°

I hope that helps!! :-)

2007-12-27 08:29:18 · answer #1 · answered by Pi R Squared 7 · 0 0

Solve it as a Quad Equation broken into Linear Equation.
Let Cosθ =x
Equation becomes
8x^2 - 2x -1 =0
Divide both sides by 8

x^2-(1/4)x-(1/8)=0

So Sum of 2 Numbers should be -1/4 and
Product of 2 Numbers should be (1)*(-1/8) = -1/8

Such two numbers are (1/2) and (-1/4)

so the equation can be written as

(x-1/2)(x-(-1/4)) = 0

Use if a*b = 0 then
either a = 0 or b = 0

similarly

x = 1/2 or x = -1/4

Cosθ = 1/2 or Cosθ =(-1/4)

You can find the θ using the inverse function.

2007-12-27 16:48:52 · answer #2 · answered by timepasse 2 · 0 0

8 cos^2 θ-2cosθ-1=0
This is a quadratic equation
let x = cosθ
the equation becomes
8x^2 - 2x -1 =0
Discriminant = (-2)^2 -4(-1)(8) =
4 +32 =36 = 6^2

x1 = (2 -6)/16 = -4/16 = -1/4
x2 = (2+6)/16 = 1/2
cosθ = -1/4 and 1/2
find θ
θ = arccos(-1/4) ----> θ = 104.47 = 104 degrees to the nearest degree.
θ = arccos(1/2) ----> θ = 60 degrees

2007-12-27 16:29:23 · answer #3 · answered by Any day 6 · 0 0

That can be factored into 2 brackets
8 cos^2 θ-2cosθ-1=0 becomes
(4cosθ+1)(2cosθ-1)=0
Looking at each case, we get
4cosθ+1=0
cosθ= -1/4

2cosθ-1=0
cosθ= 1/2

Then use a calculator to solve for each, so
θ=104.48 (~105), 60

You only have 2 solutions because your domain is only from 0 to 180 degrees

2007-12-27 16:31:17 · answer #4 · answered by ¿ /\/ 馬 ? 7 · 0 0

first thing you do is call cos(theta)=x

then you have 8x^2-2x-1=0
this is a quadratic and we can find its roots
what are the roots, using the quadratic forumlae with a=8, b=-2, and c=-1 the roots are x=1/2or -1/4,

but x=cos(theta)
so for the one root cos(theta)=1/2. when is the cosine =1/2, when x=30 degrees
for the other root cos(theta)=-1/4
so using a look up table acos(-.25)=104 degrees

2007-12-27 16:32:39 · answer #5 · answered by careyschwartz 2 · 0 0

treat this as a quadratic equation in cosx, if you solve this equation, you get cosx=1/2 or -1/4

this means that the angle is 60 degrees or 104.48 degrees

2007-12-27 16:27:23 · answer #6 · answered by kuiperbelt2003 7 · 0 0

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