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Topic:  Solving Trig. Equations 
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Subject:  reply to trig problem 
Author:  Anthony 
Date:  Nov 15 2003 
if you still need help then here it is.
the two functions you have here are inverses. So....
sec = 1/cos
csc = 1/sin
the equation then becomes
2(1/cosX)  7(1/sinX)=0
for this equation to be true 2(1/cosX) has to equal 7(1/sinX) just like
22=0 same thing.
so then
2(1/cosX)=7(1/sinX)
but then we want to get the functions on the same side.
notice that we have sin and cos if we have sin over cos then it can simplify to
tan right? yes...
so we multipy the terms first by 2 and 7 just to make it look better and get
2/cosX = 7/sinX
then multipy both sides of the equation by sinX
this produces
2sinX/cosX = 7
this simplifies to 2tanX = 7
divide by 2 and get tanX = 7/2
to get what tanX is use you calculator and use to tan^1 key that means
X=(7/2)/tan
so type (7/2) * tan^1= ?
I don't have my calculator here so you'll have to work it out from here. Good
luck
Anthony
 
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