Question 680822: Algebraically solve for x if 7 = 3 + 9sin((pi/5)(x-17))
and locate the three smallest real values of x.
I did simple algebra to get x = 17 + (5/pi)arcsin(4/9)
Then I do not know how to get the smallest real values.
The teacher says the answer is 1.267, 7.733, 11.267
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Algebraically solve for x if 7 = 3 + 9sin((pi/5)(x-17))
and locate the three smallest real values of x.
I did simple algebra to get x = 17 + (5/pi)arcsin(4/9)
Then I do not know how to get the smallest real values.
The teacher says the answer is 1.267, 7.733, 11.267
**
plot sin((π/5)(x-17)), using following coordinates:
(0,.95), (2,0), (4.5,-1), (7,0), (9.5,1), (12,0)
period of curve=2π/B=2π/(π/5)=10
7 = 3 + 9sin((π/5)(x-17))
4/9=sin((π/5)(x-17))
arcsin(4/9)≈.4606
((π/5)(x-17)=.4606
x-17=5(.4606)/π=.733
x=17.733
since the curve shifted 17 units this is not one of the 3 smallest x values
back up one period: 17.733-10=7.733
from x=7 to x=7.733 the amplitude increases 0.733
at x=2,subtract .733=1.267 to give the lowest of the 3 x-values
add one period to 1.267=1.267+10=11.267 to get the third lowest x-value
sorry, I don't know how to do it algebraically, but I thought you might want to see a graphic solution
|
|
|
