- #1
mohlam12
- 154
- 0
hey
first of all, that s not an equation, i don't 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 going to 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!
first of all, that s not an equation, i don't 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 going to 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!