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range 0 <_ x <_ 4pi
8+5sin(3x-4)=10
thanks a lot!
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)
You can shear a sheep many times but skin him only once.
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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.
Why did the vector cross the road?
It wanted to be normal.
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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)
You can shear a sheep many times but skin him only once.
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