Question

How do you find the domain of this?
\[\sqrt{ln(\cos(x))}\]

Answer 1

I sort of started it, but no clue how to put it all together.
I know that ln > 0, so x can't be 90, 270 and so on (+180k).

Answer 2

x also can't be lower than -90 i think

Answer 3

assuming everything is real

Answer 4

since `f(u) = sqrt(u)` is defined only for non negative numbers :

\(\large \ln (\cos( x)) \ge 0\)

Answer 5

\(\large \cos( x) \ge e^0\)


\(\large \cos( x) \ge 1\)

solve \(\large x\)

Answer 6

but\[\left| \cos x \right|\le 1\]
therefore \[\cos x=1,x=?\]

Answer 7

x=360k \[(k \in \mathbb{Z}^+)\]
?

Answer 8

So \[D_f=(x =2\pi k | k \in \mathbb{Z} )\]

Answer 9

looks good!

Answer 10

yay :D
thanks!

Answer 11

http://www.wolframalpha.com/input/?i=solve+over+reals+ln%28cos%28x%29%29+%3E%3D+0

Answer 12

yw:)