Trigonometric Inequality-Stuck in the end

[dolina dahani]

Hello there folks. I have a basic question of this inequality. I'm stuck in let's say end.
The Inequality is sinx-3^(1/2)*sin3x+sin5x<0 for 0-Pi
So I wrote it as equality and solved it and got to sin3x*(-(3^(1/2))+2cos2x)<0
And I got the answers for all when they are 0:
sinx=0 when x=Pi/3 and 2Pi/3
cosx=0 when x=Pi/12 and 11Pi/12
I'm not sure what to do next I mean I know how to create a table of contents but how to find - and + that's where I'm stuck. Plus is it necessary to add 2Pi? Thank you for your patience and thanks to creators of this site!

 

[Steven G]

If you write the inequality clearly you'll get help more easily.

Is it sinx-3^(1/2)*sin3x+sin5x<0 or is it sin(x-3^(1/2))*sin3x+sin5x<0?

 

[dolina dahani]

If you write the inequality clearly you'll get help more easily.

Is it sinx-3^(1/2)*sin3x+sin5x<0 or is it sin(x-3^(1/2))*sin3x+sin5x<0?

Im sorry for making it unclear. Thanks for reply!
The first one sin(x)-√3*sin(3x)+sin(5x)<0 The sin(3x) is multiplied by square root of 3

 

[Cubist]

You correctly manipulated the inequality into this form:-

[math] \sin\left(3x\right)\cdot \left(2\cos\left(2x\right)-\sqrt{3}\right) < 0[/math]
Then you meant to say:-

sin(3x)=0 when x=pi/3 and 2pi/3 (and you missed the values of x=0 and x=pi)
(2cos(2x) - sqrt(3))=0 when x=pi/12 and 11pi/12

I recommend that next you write out these values of x in increasing numerical order.

Then you need to think about what happens to the LHS at all of these points. Will the LHS value cross or just touch the x-axis at these points? IF it crosses at all points then the LHS value will change sign between every crossing point.

 

[Steven G]

I assume that you do not know calculus based on where you posted this problem.

For the record you posted your problem correctly in your 1st post--Thank You. I just wanted to be sure.

If this was my problem I would write sin(3x) and sin(5x) in terms of sin(x)

 

[dolina dahani]

You correctly manipulated the inequality into this form:-

[math] \sin\left(3x\right)\cdot \left(2\cos\left(2x\right)-\sqrt{3}\right) < 0[/math]
Then you meant to say:-

sin(3x)=0 when x=pi/3 and 2pi/3 (and you missed the values of x=0 and x=pi)
(2cos(2x) - sqrt(3))=0 when x=pi/12 and 11pi/12

I recommend that next you write out these values of x in increasing numerical order.

Then you need to think about what happens to the LHS at all of these points. Will the LHS value cross or just touch the x-axis at these points? IF it crosses at all points then the LHS value will change sign between every crossing point.

Like this???

	Pi/12	sign	pi/3	sign	2pi/3	sign	11pi/12
sin(3x)
2cos(2x)−√3
sign

 

[dolina dahani]

You correctly manipulated the inequality into this form:-

[math] \sin\left(3x\right)\cdot \left(2\cos\left(2x\right)-\sqrt{3}\right) < 0[/math]
Then you meant to say:-

sin(3x)=0 when x=pi/3 and 2pi/3 (and you missed the values of x=0 and x=pi)
(2cos(2x) - sqrt(3))=0 when x=pi/12 and 11pi/12

I recommend that next you write out these values of x in increasing numerical order.

Then you need to think about what happens to the LHS at all of these points. Will the LHS value cross or just touch the x-axis at these points? IF it crosses at all points then the LHS value will change sign between every crossing point.

I assume that you do not know calculus based on where you posted this problem.

For the record you posted your problem correctly in your 1st post--Thank You. I just wanted to be sure.

If this was my problem I would write sin(3x) and sin(5x) in terms of sin(x)

Sorry guys if I made somethings unclear my English is not the best, I dont understand all of stuff you say . The question was in Trigonometry part. But the problem is that we dont cal that part Calculus so maybe that was problem . Thank you for your patience!

 

[dolina dahani]

You correctly manipulated the inequality into this form:-

[math] \sin\left(3x\right)\cdot \left(2\cos\left(2x\right)-\sqrt{3}\right) < 0[/math]
Then you meant to say:-

sin(3x)=0 when x=pi/3 and 2pi/3 (and you missed the values of x=0 and x=pi)
(2cos(2x) - sqrt(3))=0 when x=pi/12 and 11pi/12

I recommend that next you write out these values of x in increasing numerical order.

Then you need to think about what happens to the LHS at all of these points. Will the LHS value cross or just touch the x-axis at these points? IF it crosses at all points then the LHS value will change sign between every crossing point.

Ohh sorry my English now sucks as... I understand what you said now sorry for complicating. I missed a number when i was testing......

 

[Cubist]

Like this???

	Pi/12	sign	pi/3	sign	2pi/3	sign	11pi/12
sin(3x)
2cos(2x)−√3
sign

Personally I would include the whole domain of x from 0 to pi:-

	0	< x <	pi/12	< x <	pi/3	< x <	2pi/3	< x <	11pi/12	< x <	pi
Sign of sin(3x)*(2cos(2x) - sqrt(3))		+ or -		+ or -		+ or -		+ or -		+ or -

Can you work out the signs?

 

[dolina dahani]

Personally I would include the whole domain of x from 0 to pi:-

	0	< x <	pi/12	< x <	pi/3	< x <	2pi/3	< x <	11pi/12	< x <	pi
Sign of sin(3x)*(2cos(2x) - sqrt(3))		+ or -		+ or -		+ or -		+ or -		+ or -

Can you work out the signs?

Yes I understand, thank you Sir! That was one of my subquestions on wether i should put 0and pi also, just I said 2pi as I meant in generally when it's not determined from 0 to pi. If you could just explain me why to put 0 and Pi also, is there something that I should know or is it just to make it more correct . Thank you again, and rest of you who answered

 

[Cubist]

If you could just explain me why to put 0 and Pi also, is there something that I should know or is it just to make it more correct . Thank you again, and rest of you who answered

Only because the question was " sinx-3^(1/2)*sin3x+sin5x<0 for 0-Pi "

In this question the expression is actually positive in the region from 0 -> pi/12 and therefore it does not form part of the final answer. This is probably why you are tempted to miss this region? But in the next question, or the one after, this region might form part of the answer and therefore it is a good habit to cover the whole domain in your work. Also it gives evidence to an examiner that you considered everything.

Thank you again, and rest of you who answered

You're welcome!