# Trignometric 'equation'

1. Oct 12, 2005

### mohlam12

hey
first of all, that s not an equation, i dont know the word in english (sorry)
but here... i have to solve that in the interval [-2pi , pi ]

cos(2x)-√3sin(2x) ≥ -√2

here is what i did.....
2(.5cos(2x)-√3 /2 sin(2x) ≥ -√2
cos(pi/3)cos(2x)-sin(pi/3)sin(2x) ≥ -√2 /2
cos(pi/3 + 2x) ≥ -√2 / 2
and that s gonna be....
pi/3 + 2x ≥ pi/4

x ≥ -pi/24

now what, is that all what i have to do? is waht i did right? and finally, what should the solution to this 'problem' be ????

thanks!

2. Oct 12, 2005

### mezarashi

For the record, these are called inequalities.

consider: cos z ≥ -√2 / 2
where z = pi/3 + 2x

For what values of z is this inequality satisfied? If you know the possible values of z, can you determine the possible values of x?

Last edited: Oct 12, 2005
3. Oct 12, 2005

### mohlam12

sorry, but I didnt get what you mean... All what I got from what you said is that x should be bigger or equal than (-3√2 - 2pi)/12
:huh:

4. Oct 12, 2005

### mohlam12

oh, the possible values for z are 5pi/6 , -5pi/6 , -7pi/6 i think that is it for the interval i have (-2pi , pi)
so now what ??

5. Oct 12, 2005

### mohlam12

hmmm, is that right:

the possible values of x are:
pi/6
-3pi/4
-7pi/12

are those the solutions for this INEQUALITY ??