Find all the values of x in the interval [0, 2(pi)] that satisfy the inequality: 8cos(x)+4>0.The answer should be in interval notation.

Wednesday, September 18, 2013

8*cos(x) + 4 > 0

First
we will subtract 4 from both sides.

==> 8*cos(x)
> -4

Now we will divide by
8.

$\displaystyle=\Rightarrow{\cos{{\left({x}\right)}}}>$
-4/8

$\displaystyle=\Rightarrow{\cos{{\left({x}\right)}}}>$
-1/2

$\displaystyle=\Rightarrow{\cos{{\left({x}\right)}}}+\frac{{1}}{{2}}>$
0

But we know that $\displaystyle{\cos{{\left({x}\right)}}}=-\frac{{1}}{{2}}\Leftrightarrow{x}=\frac{{{2}\pi}}{{3}},$
(4pi)/3

==> [ 0, 2pi/3 ] ==> cos(x)
> 0

==> [ 2pi/3 , 4pi/3] ==> cos(x)
< 0

==> [ 4pi/3 , 2pi] ==> cos(x)
> 0

Then, we notice that the solution
to the inequality is the interval [2pi/3 ,
4pi/3]

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