Find, to the nearest minute, the solution set of 4 cos^2 x - 9 cos x + 2 = 0 over the domain 0° less then or equal to x < 360°

Question

Find, to the nearest minute, the solution set of 4 cos^2 x - 9 cos x + 2 = 0 over the domain 0° less then or equal to x < 360°

Answer 1

4 cos²x - 9 cos x + 2 = 0

(4 cosx - 1)(cosx - 2) = 0

 

There is never a case of cosx = 2, so the solution does not exist.

 

So, we are looking for

cos x = 1/4

 

Arccos 1/4 = x

75.522488º = x  but the problem specified degrees/minutes/seconds

 

75° (0.522488)(60) 

75º 31' (.34928)(60)

75º 31' 21"

 

and  284º 28' 39"

 

Answer 2

Use this conversion factor.

 

60 minutes = 1 degree

 

 

Solve the quadratic equation that is in terms of cosx rather than x.

 

 

Use FOIL to factor.

 

(4cosx - 1)(1cosx - 2) = 0

 

 

Set the factors equal to zero.

 

4cosx - 1 = 0              and                 cosx - 2 = 0

 

cosx = 1/4                                        cosx = 2

 

 

cosx=2 has no solutions, so we reject this possibility.

 

 

 

cosx = 1/4

 

x = cos^-1(1 / 4)

 

x = 75.52

 

 

Since cosine is positive in the 1st and 4th quadrants,

 

x = 360 - 75.52

x = 284.48

 

These are your two solutions:

 

x = 75.52

x = 284.48

 

Now just use the conversion factor I have given you in the beginning to convers these solutions to minutes.

Answer 3

Hello Josh

 

4 cos² x - 9 cos x + 2 = 0

 

If we let w = cox x this gives us

 

4w²-9w+2 = 0

 

This can be factored like this.

 

(4w-1)(w-2)=0

 

Now replace the w with cos x

 

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

 

Setting each factor equal to 0,

4 cox x - 1 = 0     and      cos x - 2 = 0

4 cox x = 1                     cox x = 2

cox x = 1/4

 

The second solution, cos x = 2 is not possible since |cox x| ≤ 1

 

So we have to work with cos x = 1/4

 

We need to find the angle such that its cos is 1/4

 

Notice it is +1/4.  Cos is positive in the 1st and 4th quadrants.

 

x = 75°31'

 

Since we also need the angle in the 4th quadrant, 

 

x also = 360 - 75°31' = 284°29'

 

x = 75°31' , 284°29'

Answer 4

4cos²x - 9cos x + 2 = 0

 

Let u = cos x and substitute into the original equation:

 

     4u² - 9u + 2 = 0

     4u² - 8u - u + 2 = 0

     4u(u-2) - (u-2) = 0

     (4u-1)(u-2) = 0

 

     u = 1/4 and u = 2

 

Since u = cos x, we have cos x = 1/4 and 2.  Cos x cannot equal 2 (its range is -1 to +1), so we have:

 

     cos x = 1/4

     x = cos^-1(1/4)

     x = 75.52° and 284.48º  in the interval 0 ≤ x ≤ 360º