Question

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

Answer 1

`=> 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 )"`

Answer 2

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

Answer 3

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.