Trignometric inequality problem

1. Aug 19, 2013

needingtoknow

1. The problem statement, all variables and given/known data

Given that -pi < x < pi, solve the following inequality in radians

root(2) - 2sin(x-(pi/3)) < 0

3. The attempt at a solution

root(2) - 2sin(x-(pi/3)) < 0
- 2sin(x-(pi/3)) < -root(2)
sin(x-(pi/3)) > (root(2))

-pi<x<-pi

-pi - (pi/3) < x - pi/3 < pi - (pi/3)

-4pi < x-(pi/3) < 2pi/3

and I know that -5pi/4 and pi/4 fit the new range as a result of the translation pi/3 I just don't know what to do with them. Can someone please help?

That is so far what I have got I do not know where to proceed from

2. Aug 19, 2013

Staff: Mentor

You have a mistake in the 3rd line. You can go from the first inequality directly to this inequality by adding 2sin(x - $\pi/3$) to both sides.

$\sqrt{2} < 2sin(x - \pi/3)$
??? This is saying that $-\pi$ is less than itself, which is not true. How did you get this?

3. Aug 19, 2013

needingtoknow

oh sorry thats a typo its actually -pi < x < +pi and root(2)/2

4. Aug 19, 2013

Staff: Mentor

Can you solve sin(u) > $\sqrt{2}/2$? There are lots of intervals that satisfy this inequality. Once you get that, then solve for x, where u = x - $\pi/3$.

5. Aug 19, 2013

needingtoknow

Is there a way to solve this without the use of a graph though?

6. Aug 19, 2013

Staff: Mentor

I suppose, but why would you not want to use a graph? That's probably the simplest way to proceed.