How to solve $2 {cos}^{2} x + 5 cos x + 2 \geq 0$ $2cos^2x+5cosx+2>=0$ ?

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How to solve $2 {cos}^{2} x + 5 cos x + 2 \geq 0$ $2cos^2x+5cosx+2>=0$ ?

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Answer 1 · 2018-06-07T11:45:13

$\Rightarrow x \in [2 k π, \frac{2 π}{3} + 2 k π] \cup [\frac{4 π}{3} + 2 k π, 2 π (k + 1)]$ $=> x in [2k pi, (2 pi)/3 + 2 k pi] uu [(4 pi)/3 + 2k pi, 2 pi (k+1)]$
$(k any integer )$ $"( "k" any integer )"$

Explanation:

$Name y = cos(x)$ $"Name y = cos(x)"$

$2 y^{2} + 5 y + 2 \geq 0$ $2 y^2 + 5 y + 2 >= 0$

$Now solve$ $"Now solve"$

$2 y^{2} + 5 y + 2 = 0$ $2 y^2 + 5 y + 2 = 0$

$Discriminant : 5^{2} - 4 \cdot 2 \cdot 2 = 9 = 3^{2}$ $"Discriminant : "5^2 - 4*2*2 = 9 = 3^2$
$\Rightarrow y = \frac{- 5 \pm 3}{4} = - 2 or - \frac{1}{2}$ $=> y = (-5 pm 3)/4 = -2 or -1/2$

$Between those 2 roots 2 y^{2} + 5 y + 2 < 0 .$ $"Between those 2 roots "2 y^2 + 5 y + 2 < 0.$
$\Rightarrow 2 y^{2} + 5 y + 2 \geq 0 if y \leq - 2 or y \geq - \frac{1}{2} .$ $=> 2 y^2 + 5 y + 2 >= 0 " if "y<=-2" or "y>=-1/2.$

$\Rightarrow cos (x) \leq - 2 (this is impossible as - 1 \leq cos (x) \leq 1)$ $=> cos(x) <= -2" (this is impossible as "-1<=cos(x)<=1")"$
$or$ $"or"$
$cos (x) \geq - \frac{1}{2}$ $cos(x)>=-1/2$
$\Rightarrow x \in [2 k π, \frac{2 π}{3} + 2 k π] \cup [\frac{4 π}{3} + 2 k π, 2 π (k + 1)]$ $=> x in [2k pi, (2 pi)/3 + 2 k pi] uu [(4 pi)/3 + 2k pi, 2 pi (k+1)]$
$(k any integer )$ $"( "k" any integer )"$

Answer 2 · 2018-06-07T12:09:41

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 $4 {cos}^{2} x$ $4cos^2x$ and the sum is $5 cos x$ $5cosx$ so the factors are $cos x$ $cosx$ and $4 cos x$ $4cosx$
$2 {cos}^{2} x + cos x + 4 cos x + 2 \geq 0$ $2cos^2x+cosx+4cosx+2>=0$
we then factorise (ignore the inequality for now)
$cos x (2 cos x + 1) + 2 (2 cos x + 1) = 0$ $cosx(2cosx+1)+2(2cosx+1)=0$
$(2 cos x + 1) (cos x + 2) = 0$ $(2cosx+1)(cosx+2)=0$
and solve

$2 cos x = - 1$ $2cosx=-1$
$cos x = - \frac{1}{2}$ $cosx=-1/2$
$x = arccos (- \frac{1}{2})$ $x=arccos(-1/2)$

From your exact ratios you should know that $cos x = \frac{π}{3}$ $cosx=pi/3$ and cos is negative in the 2nd and 3rd quadrants. So $x = - 2 \frac{π}{3} and 2 \frac{π}{3} and 4 \frac{π}{3}$ $x=-2pi/3 and 2pi/3 and 4pi/3$ 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 · 2018-06-07T15:47:00

Half closed intervals $(0, \frac{2 π}{3}] and [\frac{4 π}{3}, 2 π)$ $(0, (2pi)/3] and [(4pi)/3, 2pi)$

Explanation:

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

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How to solve $2 {cos}^{2} x + 5 cos x + 2 \geq 0$ $2cos^2x+5cosx+2>=0$ ?

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