Asked by paul
Hello there,
im doing an equation and its mind boggling me badly so i want to see if im doing something wrong or not.
It goes like this:
sinx - Squareroot of 3/2 > 0
x € [0,2pi)
i start by moving the sqroot of 3.. so i have sinx >sq3/2.
as of that my resaults are 1,05 + n*2pi, -1,05 + n*2pi , 2,09+n*2pi and 4,19+n*2pi. (i get -1,05 and 4,19 since we are taking the sqroot of 3 so we get plus and minus answere?
I hope im doing something right
b.r.g.
Paul
im doing an equation and its mind boggling me badly so i want to see if im doing something wrong or not.
It goes like this:
sinx - Squareroot of 3/2 > 0
x € [0,2pi)
i start by moving the sqroot of 3.. so i have sinx >sq3/2.
as of that my resaults are 1,05 + n*2pi, -1,05 + n*2pi , 2,09+n*2pi and 4,19+n*2pi. (i get -1,05 and 4,19 since we are taking the sqroot of 3 so we get plus and minus answere?
I hope im doing something right
b.r.g.
Paul
Answers
Answered by
Steve
sinx - √3/2 > 0
sinx > √3/2
you know that sin π/3 = √3/2
and sinx > 0 for 0<x<π
and sin(π-x) = sin(x)
So, sinx > √3/2 for π/3 < x < 2π/3
Maybe a little look-see at the graph of sin(x) would help your intuition.
http://www.wolframalpha.com/input/?i=sin%28x%29+%3E+%E2%88%9A3%2F2+for+0+%3C+x+%3C+2pi
sinx > √3/2
you know that sin π/3 = √3/2
and sinx > 0 for 0<x<π
and sin(π-x) = sin(x)
So, sinx > √3/2 for π/3 < x < 2π/3
Maybe a little look-see at the graph of sin(x) would help your intuition.
http://www.wolframalpha.com/input/?i=sin%28x%29+%3E+%E2%88%9A3%2F2+for+0+%3C+x+%3C+2pi