(cos x)^2 -1.2cos x-0.28=0?

Question

My question is to solve

(cosx)^2 -1.2cosx-0.28=0

for "x" in the interval [0, 2pi].

I have gotten to a point, but seem to be stuck. I have done the quadratic formula and got the answers of:

cos x = 1.4 and cos x=-0.2

I then use the inverse functions and get

x=1.77215...
(I can not take the arc cos of 1.4)
but now I am stuck.

The question says to use the quadratic formula, inverse trig functions, and the symmetry of the unit circle.

There must be a step I am missing with the last part.
Can anyone point out to me what that is?
Thanks!

Answer 1

[cos(x)]^2 - 1.2cos(x) - 0.28 = 0

Multiply through by 100 to get rid of decimals.
100[cos(x)]^2 - 120cos(x) - 28 = 0

Divide by the common factor = 4.
25[cos(x)]^2 - 30cos(x) - 7 = 0

This expression factors, so we don't need the quadratic formula.
[ 5cos(x) - 7 ] [ 5cos(x) + 1 ] = 0

Equating each factor to zero gives :
cos(x) = 7/5 = 1.4, but cos(x) is always <= 1,
so this is not a solution.

or

cos(x) = -1/5 = - 0.2
This is a solution because -1 <= cos(x) <= 1.

Taking the arccos by calculator gives :
x = 101.54º

This angle is in the second quadrant.

For cos, there must also be an angle in the third quadrant,
because the cos of all angles there, are also negative.

This angle is greater than 180º by the same
amount that the other angle is less than 180º.

So we find that 180º - 101.54º = 78.46º
Thus, 101.54º is less than 180º by 78.46º.

Therefore, the other angle must be 78.46º more than 180º.
So the answer to this is 180º + 78.46º = 258.46º.

Therefore, x = 101.54º or 258.46º.

I'll let you calculate those values in radians.
I still find it easier to work with degrees!!!

Answer 2

Hi,

The reason that you can't solve for arccos 1.4 is that the domain of arccosine is (-1, 1).

That's fine, just use arccos (-0.2) and then since cos is symmetric about the x axis, if you add (pi/4) to your solution you get another answer.

example
arccos (1/2) = pi/4 and -pi/4

Hope that helps,
Matt

Answer 3

(cos x)^2 -1.2 cosx -.28 = 0
Let z=cos x
Then z^2 -1.2 z -.28 = 0
z=1.2 +- sqr[(1.2)^2 -4(1)(-.28)] = 1.2 +- sqr[1.44+ 1.12]
=1.2+- sqr[2.56] = cos x
So, whatever sqr(2.56) is, the value of cos x is not greater than one, so only the - sqr makes sense. And the exact answer is
cos x = 1.2 - sqr(2.56), or
x = arccos[1.2 - sqr(2.56)]