Findind root of sin x

[learningmathh]

hello everyone,

I want to find root for \(\displaystyle f(x)=2-3sin(\frac{\pi }{4}x-\frac{\pi}{2}), x\epsilon [0,8]\)

I reached so far \(\displaystyle sin\frac{\pi}{4}x+\frac{\pi}{2}=\frac{2}{3}\)

Pleas help, I have used alot of enery in this one.

 

[Deleted member 4993]

hello everyone,

I want to find root for \(\displaystyle f(x)2-3sin(\frac{\pi }{4}x-\frac{\pi}{2}), x\epsilon [0,8]\)

I reached so far \(\displaystyle sin\frac{\pi}{4}x=\frac{2}{3}+\frac{\pi}{2}\)

Pleas help, I have used alot of enery in this one.

root of what?

Is your problem:

f(x) = 2 - 3*sin(\(\displaystyle \frac{\pi}{4}x \ - \ \frac{\pi}{2}\))

If it is then the work you have shown - tells me that you need to review the definition "roots of functions". Also you need to edit your post and correct it.

Then show your work again.

 

[HallsofIvy]

hello everyone,

I want to find root for \(\displaystyle f(x)2-3sin(\frac{\pi }{4}x-\frac{\pi}{2}), x\epsilon [0,8]\)

I reached so far \(\displaystyle sin\frac{\pi}{4}x=\frac{2}{3}+\frac{\pi}{2}\)

This, to begin with is wrong. If 2- 3\sin(\frac\left{\pi}{4}x- \frac{\pi}{2}\right)= 0
(Strictly speaking an equation has roots. I assume you are looking for the zeroes of f(x), the roots of f(x)= 0.)
then, first subtracting 2 from both sides and dividing both sides by -3, you have \(\displaystyle \sin\left(\frac{\pi}{4}x\right)= \frac{2}{3}\).
But you cannot take the "\(\displaystyle \frac{\pi}{2}" out of the sine function like that- take the inverse sine (or arcsine) first, then add \(\displaystyle \frac{\pi}{2}\) to both sides.

Pleas help, I have used alot of enery in this one.

\)

 

[learningmathh]

root of what?

Is your problem:

f(x) = 2 - 3*sin(\(\displaystyle \frac{\pi}{4}x \ - \ \frac{\pi}{2}\))

If it is then the work you have shown - tells me that you need to review the definition "roots of functions". Also you need to edit your post and correct it.

Then show your work again.

Yes that is my probem. I forgot =, sorry.

 

[learningmathh]

This, to begin with is wrong. If 2- 3\sin(\frac\left{\pi}{4}x- \frac{\pi}{2}\right)= 0
(Strictly speaking an equation has roots. I assume you are looking for the zeroes of f(x), the roots of f(x)= 0.)
then, first subtracting 2 from both sides and dividing both sides by -3, you have \(\displaystyle \sin\left(\frac{\pi}{4}x\right)= \frac{2}{3}\).
But you cannot take the "\(\displaystyle \frac{\pi}{2}" out of the sine function like that- take the inverse sine (or arcsine) first, then add \(\displaystyle \frac{\pi}{2}\) to both sides.\)

\(\displaystyle

Yes, my mathenglish needs improving.

You are right. I did as you said, thanks for the help \)

 

[HallsofIvy]

hello everyone,

I want to find root for \(\displaystyle f(x)= 2-3sin(\frac{\pi }{4}x-\frac{\pi}{2}), x\epsilon [0,8]\)

I reached so far \(\displaystyle sin\frac{\pi}{4}x=\frac{2}{3}+\frac{\pi}{2}\)

This, to begin with, is wrong. If 2- 3\sin(\frac\left{\pi}{4}x- \frac{\pi}{2}\right)= 0[/tex]
(Strictly speaking an equation has roots. I assume you are looking for the zeroes of f(x), the roots of f(x)= 0.)
then, first subtracting 2 from both sides and dividing both sides by -3, you have \(\displaystyle \sin\left(\frac{\pi}{4}x\right)= \frac{2}{3}\).
But you cannot take the "\(\displaystyle \frac{\pi}{2}\)" out of the sine function like that- take the inverse sine (or arcsine) first, then add \(\displaystyle \frac{\pi}{2}\) to both sides.

Pleas help, I have used alot of enery in this one.