反tan函數解出x?

令 :
a = tan-1(5x) , b = tan-1[ 4x/(6-x^2) ] , c = tan-1(x)
故得:
a + b - c = 90°
5x = tan a , 4x/(6-x^2) = tan b , x = tan c

tan ( a - c ) = tan ( 90° - b ) = cot b
( tan a - tan c ) / ( 1 - tan a*tan c ) = 1 / tan b
( 5x - x ) / ( 1 - 5x*x ) = ( 6 - x^2 ) / ( 4x )
5x^4 - 13x^2 - 6 = 0
( x^2 - 3 )( 5x^2 + 2 ) = 0
x^2 = 3 , - 0.4 (不合)
x = ± √3
這兩個根代回原式皆成立

Ans: x = ± √3