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Does anyone know how to solve [1+tan(2)x / tan(2)x]=csc(2)x?
3 Answers
 oregfiuLv 71 decade agoFavorite Answer
The equation is not quite clear. I suppose that you mean this:
[1 + tan(2x)] / [tan(2x)] = csc(2x)
If yes, then
cotg(2x) + 1 = csc(2x)
cos(2x) + sin(2x) = 1
(√2) sin(2x + π/4) = 1
sin(2x + π/4) = √2/2
2x + π/4 = arcsin(√2/2)
Solutions are
x = πn
and
x = π/4 + πn
where n is any whole number.

Note: To square both sides of an equation is not a correct way. It gives additional false results.

 ted sLv 71 decade ago
If the (2) was to mean squared then your statement is an identity, cot^2( x )+ 1 =
csc^2 (x). If it meant 2x then it can be written as cos x + sin x = 1. Square both sides ,clean it up, you find sin(4x) = 0 which has answers n(pi / 4) , n = 0,1,2,3,4,5,6,7 for answers in [0,2 pi)
 LE THANH TAMLv 51 decade ago
[ 1 + tan(2)x / tan(2)x ] = csc(2)x
we have : tan2x = sin2x / cos2x
we have
1 +tan2x =( csc2x)(tan2x) = (1 /sin2x)(sin2x / (cos2x)
1 +tan2x = 1 / cos2x
cos2x( 1 +tan2x ) = 1
cos2x + (tan2x) (cos2x) = 1
cos2x +[( sin2x / cos2x)(cos2x)] = 1
cos2x + sin2x = 1
we have ( csc2x = ( 1 / sin2x) =( csc2x) ( sin2x) = 1
we have
cos2x + sin2x =( csc2x) (sin2x)
we have
csc2x = [( cos2x + sin2x ) / sin2x ]
= (cos2x / sin2x) + (sin2x /sin2x)
but ( cot (2x )= ( cos2x / sin2x)
csc2x = cot 2x + 1
but ; cot 2x = 1 / tan2x
csc2x = ( 1 / tan2x) + 1
csc2x = (1 + tan2x) / tan2x =
(1 + tan2x ) / tan2x = csc2x