cos(sqrt(x))=0 i got that x = (2n+1)*pi/2 where n = integers from 1 to infinity. is this correct?
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Originally Posted by gsds cos(sqrt(x))=0 i got that x = (2n+1)*pi/2 where n = integers from 1 to infinity. is this correct? $\displaystyle \sqrt{x} = (2n+1)\left(\frac{\pi}{2}\right)$ , $\displaystyle n \in \mathbb{Z}$ square both sides.
Originally Posted by gsds cos(sqrt(x))=0 i got that x = (2n+1)*pi/2 where n = integers from 1 to infinity. is this correct? cos(sqrt(x))=0 sqrt(x)= (2n+1)*pi/2 x = [(2n+1)*pi/2]^2 (might want to distribute this out)
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