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solve for x: sin ((pi/4)x^2) = 1
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first solve: sin(u)=1 u=pi/2 but we could keep going a round and a round the circle so u=pi/2+2npi n is an integer so if sin(u)=1 when u=pi/2+2npi then sin((pi/4)x^2)=1 when (pi/4)x^2=pi/2+2npi
solve that last equation for x and woolah
ick
im confused! how do i solve that last eqn?
using algebra multiply by 4/pi on both sides then take the square root of both sides don't forget plus or minus
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is "woolah" arkansian for voila?
lol
its my new word
too much barbecue and beer for me.
i will use it next week
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you can use it i guess
lol whats the answer? :P
i will say it like inspector clouseau
(sqrt) 2 + 8n ???
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