How to solve 2cos^2x+5cosx+2>=0 ?

3 Answers
Jun 7, 2018

=> x in [2k pi, (2 pi)/3 + 2 k pi] uu [(4 pi)/3 + 2k pi, 2 pi (k+1)]
"( "k" any integer )"

Explanation:

"Name y = cos(x)"

2 y^2 + 5 y + 2 >= 0

"Now solve"

2 y^2 + 5 y + 2 = 0

"Discriminant : "5^2 - 4*2*2 = 9 = 3^2
=> y = (-5 pm 3)/4 = -2 or -1/2

"Between those 2 roots "2 y^2 + 5 y + 2 < 0.
=> 2 y^2 + 5 y + 2 >= 0 " if "y<=-2" or "y>=-1/2.

=> cos(x) <= -2" (this is impossible as "-1<=cos(x)<=1")"
"or"
cos(x)>=-1/2
=> x in [2k pi, (2 pi)/3 + 2 k pi] uu [(4 pi)/3 + 2k pi, 2 pi (k+1)]
"( "k" any integer )"

I'm a little baffled since arccosx=-2 doesnt exist but id already typed half this up so i figured i may as well post it. If you can find a mistake i wouldnt be suprised.

Explanation:

As with a general inequality algebra question my response would be: factorise, solve for equal to x (instead of the inequality) and then investigate at the key points.
So this thing factorises like a nice quadratic, the product is 4cos^2x and the sum is 5cosx so the factors are cosx and 4cosx
2cos^2x+cosx+4cosx+2>=0
we then factorise (ignore the inequality for now)
cosx(2cosx+1)+2(2cosx+1)=0
(2cosx+1)(cosx+2)=0
and solve

2cosx=-1
cosx=-1/2
x=arccos(-1/2)

From your exact ratios you should know that cosx=pi/3 and cos is negative in the 2nd and 3rd quadrants. So x=-2pi/3 and 2pi/3 and 4pi/3 etc.

solving the other part gives us
x=arccos(-2) which is bad and we broke it.
im not a math expert and i no longer know what to do
I was then going to test a number in between each of these intervals and see which ones gave me positive and negative results but idk anymore

Jun 7, 2018

Half closed intervals (0, (2pi)/3] and [(4pi)/3, 2pi)

Explanation:

First find the end-points (critical points) by solving the quadratic equation for cos x:
f(x) = 2cos^2 x + 5cos x + 2 = 0
D = d^2 - 4ac = 25 - 16 = 9 --> d = +- 3
There are 2 real roots:
cos x = -b/(2a) +- d/(2a) = - 5/4 +- 3/4.
cos x = -2 (rejected), and cos x = - 1/2 .
Trig table and unit circle give 2 end-points:
x = +- 2pi/3, or x = (2pi)/3, and x = (4pi)/3 (co-terminal)
The graph of f(x) is an upward parabola --> f(x) > 0 when x is out side the 2 real roots --> cos x <= - 2 (rejected), and
- 1/2 <= cos x
By considering an arc x that rotates counterclockwise on the unit circle, we see that cos x >= - 1/2 when x varies inside the 2 half closed intervals (0, (2pi)/3], and [(4pi)/3, 2pi) that are the answers.
For general answers , just add 2kpi.
Check.
x = pi/2 --> 2cos^2 x = 0 --> 5cos x = 0 --> f(x) = 2 > 0. Proved
x = pi --> 2cos^2 x = 2 --> 5cos x = - 5 -->
f(x) = 2 - 5 + 2 = - 1 < 0. Proved.