How to solve #2cos^2x+5cosx+2>=0# ?
3 Answers
Explanation:
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
we then factorise (ignore the inequality for now)
and solve
From your exact ratios you should know that
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
Half closed intervals
Explanation:
First find the end-points (critical points) by solving the quadratic equation for cos x:
There are 2 real roots:
cos x = -2 (rejected), and
Trig table and unit circle give 2 end-points:
The graph of f(x) is an upward parabola --> f(x) > 0 when x is out side the 2 real roots -->
By considering an arc x that rotates counterclockwise on the unit circle, we see that
For general answers , just add
Check.
f(x) = 2 - 5 + 2 = - 1 < 0. Proved.