Question

how to find csc(x)=sec(x) for x from 0,2pi

Answer 1

cosx=sinx <-cosx=sin(x+pi/2)
sin(x+pi/2)=sinx
.: x+pi/2=x ??

Answer 2

x-x=-pi/2
0=-pi/2, haha.. lol nice joke^^

Answer 3

i think sin(x) = cos(x) if x=pi/4 and x=5pi/4

Answer 4

ye but why cant you do what i did

Answer 5

U have Tan x = 1 so in the range 0 to 2pi you have x=?

Answer 6

yeah why cant do my way though ??

Answer 7

i mean whats wrong with what i did

Answer 8

you are missing solutions, and you have made a mistake in the identity

Answer 9

\[\cos(x)=\sin(x)\]
\[\sin(x-\pi/2)=\sin(x)\]
the arc sine is not defined for both sides at the same time

Answer 10

should be sin( pi/2 - x) = sin(x)

Answer 11

so was the problem when i subbed sin(pi/2+x)=sinx

Answer 12

i thought sin(pi/2+x)=sinx as well

Answer 13

cosx*

Answer 14

ah yes. i think the problem is Arcsin

Answer 15

Hello - so you've got sin x = cos x . Where are the angles of such S-Y-M-M-E-T-R-Y ?

Answer 16

?? so was my substitution of sin(pi/2+x) invalid

Answer 17

taking the arcsin of both sides was the invalid step