#ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#是否需先證(8x^3-6x+1)=0之三根皆在-1~1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#,才可令 x = s in@#ed_op#/DIV#ed_cl#galaxylee 寫到:#ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#16x^4-8x^3-12x^2+8x-1=0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(2x-1)(8x^3-6x+1)=0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#x=1/2是一根#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#8x^3-6x+1=0 => 2(3x-4x^3)=1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#令x=sinθ 上式變成 2sin3θ=1 => sin3θ=1/2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#3θ=(π/6)+2kπ,或(5π/6)+2kπ,其中k為整數#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#θ=(π/18)+(2kπ)/3,或(5π/18)+(2kπ)/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#可歸納出sinθ=sin(π/18) 或 sin(5π/18),sin(-7π/18)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以方程式的四根為 x = 1/2(=sin(π/6))、sin(π/18)、sin(5π/18)、sin(-7π/18)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#故 sin^(-1)x = π/6、π/18、5π/18、-7π/18#ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#