Trignometric inequality problem

[needingtoknow]

Homework Statement

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

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

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

 

[Mark44]

Homework Statement

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

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

The Attempt at a Solution

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

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)##

-pi<x<-pi

? This is saying that ##-\pi## is less than itself, which is not true. How did you get this?

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

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

 

[needingtoknow]

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

 

[Mark44]

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##.

 

[needingtoknow]

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

 

[Mark44]

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

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