joachimngeny
  • joachimngeny
can someone help with this. find the smallest solution, the largest solution and the number of solutions for x in the interval 0 ≤ x ≤ 2π
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
joachimngeny
  • joachimngeny
im stuck with the second solution
madhu.mukherjee.946
  • madhu.mukherjee.946
am i wrong
joachimngeny
  • joachimngeny
@madhu.mukherjee.946 i just realised i didnt attach the equation sin(7 x) = 0.64

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

madhu.mukherjee.946
  • madhu.mukherjee.946
ok now its f9
joachimngeny
  • joachimngeny
@Michele_Laino i just realise that i have to give answer in radian
Michele_Laino
  • Michele_Laino
hint: |dw:1440747531526:dw|
madhu.mukherjee.946
  • madhu.mukherjee.946
did you get the value of x
Michele_Laino
  • Michele_Laino
the smallest solution (in radians) is: \[\Large {x_1} = \frac{1}{7}\arcsin \left( {0.64} \right) = 0.0992\]
joachimngeny
  • joachimngeny
i got x1 = 0.10 and X2= 3.04
Michele_Laino
  • Michele_Laino
whereas the second one, is: \[\Large {x_2} = \pi - 0.0992 = ...?\]
Michele_Laino
  • Michele_Laino
I have used windows calculator and the "radians" option
Michele_Laino
  • Michele_Laino
your answer is correct!
Michele_Laino
  • Michele_Laino
since you have used approximated values
joachimngeny
  • joachimngeny
@Michele_Laino apparently my answer is partially wrong.
joachimngeny
  • joachimngeny
I must have rounded it off wrong or something
Michele_Laino
  • Michele_Laino
yes! you have rounded off the smallest solution, nevertheless it is not a wrong answer
joachimngeny
  • joachimngeny
@Michele_Laino could you show me how to get it right, using radian? Becuase i found the angles in degree then i changed it to radians that probably why im losing marks
Michele_Laino
  • Michele_Laino
here is the right setting on windows calculator
joachimngeny
  • joachimngeny
@Michele_Laino is that what you got for the smallest value?
joachimngeny
  • joachimngeny
im on mac computer atm
Michele_Laino
  • Michele_Laino
no, it is the smallest value multiplied by 7, namely: 7*x1=0.6944
Michele_Laino
  • Michele_Laino
as you can see from my image, there is the formula: "asinr(0.64)" at the upper right corner
Michele_Laino
  • Michele_Laino
now, in order to get the value of x1, you have to divide 0.694498... by 7, so you get this: 0.0992...
joachimngeny
  • joachimngeny
yes i understand. this is what i answered
joachimngeny
  • joachimngeny
@Michele_Laino
Michele_Laino
  • Michele_Laino
I'm reading...
joachimngeny
  • joachimngeny
what i did seem right, yeah?
Michele_Laino
  • Michele_Laino
please wait a moment, I have to answer to my phone...
Michele_Laino
  • Michele_Laino
ok! I'm here
joachimngeny
  • joachimngeny
@Michele_Laino did you see my previous reply
Michele_Laino
  • Michele_Laino
yes! I see, I'm sorry the second solution is: \[\Large 7{x_2} = \pi - 7{x_1} = \pi - \left( {7 \cdot 0.0992} \right) = ...?\] because, from my drawing above I get these values: \[\Large 7{x_1},7{x_2}\]
Michele_Laino
  • Michele_Laino
therefore: \[\Large \begin{gathered} {x_1} = \frac{1}{7}\arcsin \left( {0.64} \right) = 0.0992, \hfill \\ \hfill \\ {x_2} = \frac{{\pi - \left( {7 \cdot 0.0992} \right)}}{7} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
number of solution is correct, namely, we have 14 solutions
Michele_Laino
  • Michele_Laino
and the highest solution is: \[\Large {x_2} + 6 \cdot \frac{{2\pi }}{7}\]
joachimngeny
  • joachimngeny
I don't know why it says partially wrong.
joachimngeny
  • joachimngeny
@Michele_Laino but isnt that outside the range? 0 -> 2pi?
Michele_Laino
  • Michele_Laino
I think that tne first value, namely 0.1 is right, the third value, namely 14 is also right, you have to correct the second value, namely: \[\Large {x_2} + 6 \cdot \frac{{2\pi }}{7} = \frac{{\pi - \left( {7 \cdot 0.0992} \right)}}{7} + 6 \cdot \frac{{2\pi }}{7} = 5.735\]
Michele_Laino
  • Michele_Laino
namely 3.04 is wrong, please replace it with 5.735
Michele_Laino
  • Michele_Laino
better is 5.74
joachimngeny
  • joachimngeny
@Michele_Laino ok thank you. I appriciate your help
Michele_Laino
  • Michele_Laino
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.