How do you solve #tan^-1(2x)+tan^-1(x)= (3pi)/17#?
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#tan^-1 (2x) #--> arc (2x)
#tan ^-1 x # --> arc x
The equation is transformed to:
#2x + x = 3x = (3pi)/17#
#x = pi/17#
Answers in degrees:
#pi/17 = 10^@59#
#(2pi)/17 = 21^@18#
Let #c = tan ((3pi)/17) = 0.619174#, nearly. Then, #x =(-3+- sqrt(9+8c^2))/(4c)=0.19129 and -2.61367#, nearly It is seemingly Incredible that x could be negative.
Let #a = tan^(-1)(2x) and b = tan^(-1) (x)#.
Then #tan (a + b ) = (tan a + tan b )/(1-tna tan b )#
#=(2x+x)/(1-(2x)(x))#
#=(3x)/(1-2x^2) = tan ((3pi)/17) = c=0.619174#, nearly.
Solving this quadratic for a positive x,
#x = (sqrt(9+8c^2)-3)/(4c)=0.19129#, nearly.
Here, the two angles that add up to #(3pi)/17 =31.76^o# are #20.94^o
and 10.83^o#.
Seemingly incredible but true, the negative root #=-2.61367# is
also a solution.
The corresponding angles are #100.83^o and -69.06^o#.
I have used that tangent is negative in the 2nd quadrant for
#tan^(-1)(2x)# and in the fourth quadrant for #tan^(-1)(x)#, respectively.
