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
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
Hello Paul,
Let's go through the steps together to see if you're on the right track.
To solve the inequality sinx - sqrt(3/2) > 0, we can start by moving sqrt(3/2) to the other side of the inequality, so we have sinx > sqrt(3/2).
Now, let's consider the possible values of sinx in the given interval [0, 2pi).
In the first quadrant (0 to pi/2), sinx is positive, so it satisfies the inequality.
In the second quadrant (pi/2 to pi), sinx is also positive.
Now, in the third quadrant (pi to 3pi/2), sinx is negative, and it does not satisfy the inequality.
Finally, in the fourth quadrant (3pi/2 to 2pi), sinx is negative, but it satisfies the inequality because we're considering x in the interval [0, 2pi) and 2pi is not included.
So, the values of x that satisfy the inequality are in the intervals [0, pi/2) and (3pi/2, 2pi).
To summarize, your conclusion is correct. The solutions for x in the given interval [0, 2pi) are x = 1.05 + n*2pi, where n is an integer.
I hope this helps! Let me know if you have any further questions.