Question

Find exact solutions of the equation in the interval [0,pi) cos y cot^2 y = cos y

Answer 1

Help, please! :)

Answer 2

@Miracrown

Answer 3

Anybody have any ideas? o.o

Answer 4

\[\huge\rm \rm cos y \cot^2 y = \cos y \]
divide both side by cos y
.....?

Answer 5

so we're left with

cot^2y
------
cosy

Answer 6

\[\huge\rm \frac{ \cancel{cos y} \cot^2 y }{\cancel{cos y}}= \frac{\cancel{ \cos y}}{\cancel{cos y}} \]
\[\large\rm cot^2 y = 0\]

Answer 7

Division is badddd +_+
We lose some stuffffff
I think subtracting cosy to the left side is probably betterrrrr :D

Answer 8

\[\Large\rm \cos y \cot^2y-\cos y=0\]:3

Answer 9

=.= okayyyyyyyyyyyyyyy
then subtaraarrararaaaact
lololol

Answer 10

yeah that's easy ;3

Answer 11

you guys haha ok now what o.0

Answer 12

nesh, your pictures are better lately 0_o
less creepy sad babies = good

Answer 13

now factor out a cosy from each term :)

Answer 14

which would look like...

Answer 15

cosy (cot^2-1) = 0 ?

Answer 16

Good c: then apply your `zero factor property`:
`cosy=0`
and
`cot^2y-1=0`

Answer 17

so then the exact solutions are what?

Answer 18

uhh.. well.. solve for them separately.
cos y = 0

what angles does this correspond to?
when cosine is giving you 0, which angles? :o

Answer 19

\[\huge\rm (x , y ) \rightarrow (\cos , \sin ) \]
;3

Answer 20

er 0? haha

Answer 21

oh great the babies are back -_-

cosine is 0 at 0 degrees? no.

Answer 22

cosine of 0 degrees is 1, yes?

Answer 23

so that's no bueno

Answer 24

pi/2

Answer 25

lO_Ok at the unit circle o^_^o