1. ## trignometric solution

cos(sqrt(x))=0

i got that x = (2n+1)*pi/2 where n = integers from 1 to infinity.

is this correct?

2. 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?
$\sqrt{x} = (2n+1)\left(\frac{\pi}{2}\right)$ , $n \in \mathbb{Z}$

square both sides.

3. 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)