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Neela

Guest

I need help solving this equation

range 0 <_ x <_ 4pi

8+5sin(3x-4)=10

thanks a lot!

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: I need help solving this equation

range 0 <_ x <_ 4pi

8+5sin(3x-4)=10

thanks a lot!

subtracting 8 from each side:

5sin(3x-4)=2

dividing both sides by 5:

sin(3x-4)=2/5

rewriting:

arcsin(2/5)=3x-4

adding 4 to both sides:

arcsin(2/5)+4=3x

dividing both sides by 3:

[arcsin(2/5)+4]/3=x

x=1.4705 (approximately)

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: I need help solving this equation

What you've put is correct, but it's not complete.

arcsin(2/5) is given as ≈ 0.411 when you put it into a calculator, but in fact it takes many values.

It can also be equal to ≈ 2.730 (because that is π - 0.411) and also also either of those two numbers ± 2nπ, where n is any integer.

The range in the question is given as 0<x<4π, so for a complete answer you need to find all values of arcsin(2/5) that give an x within that range.

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It wanted to be normal.

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: I need help solving this equation

What you've put is correct, but it's not complete.

True. There are 11 other possible (approximate) values of x, although the exact answer I listed actually includes all of them.

Last edited by All_Is_Number (2006-11-25 09:10:18)

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You can shear a sheep many times but skin him only once.