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Aug 9th, 2002, 12:09 PM
#3
Lively Member
!
hmhm maybe this :
8^[1-cos^2(x)]+8^[cos^2(x)]=10+cos2y
=>
8
----------------- + 8^[cos^2(x)]=10+cos2y
8^[cos^2(x)]
let A=8^[cos^2(x)]
now:
8
----- + A =10+cos2y
A
8+ A^2
----------- = 10+cos2y
A
im going to prove this:
8+A^2
--------- <=9
A
=> A^2 - 9A +8 <=0
A1,2=[9+_sqrt(81-32)]/2
A1,2=8,1 (*)
+++ ;;;;;;;;;;;;;;;; ++++++++
------1---------------8-------------
;;;;;; ---------------- ;;;;;;;;;;;;;
we have now that
8+A^2
---------- <=9 (**)
A
and (|cos2y|<1) => 9<=(10-cos2y)<=11 (***)
using (**) & (***)
left side is equal to right only if both sides is equal to 9
now using (*) A1,2 = 1,8
(I) A=8^[cos^2(x)]=1
cos^2(x)=0 => {{{{ cos(x)=0 }}}}
10+cos2y=9 => {{{{ cos2y=1 }}}}
(II) A=8^[cos^2(x)]=8
cos^2(x)=1
10+cos2y=9
{{{{ cos(x)=1 }}}} or {{{{ cos(x)=-1 }}}}
{{{{ cos2y =1 }}}} {{{{ cos2y =1 }}}}
finaly:
(i) x=PI/2+k*PI ; k={0,1,........}
y=k*PI ; k={0,1,........}
(ii)x=2*k*PI ; k={0,1,........}
y=k*PI ; k={0,1,........}
(iii)x=PI+2*k*PI ; k={0,1,........}
y=k*PI ; k={0,1,........}
SEE YA
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