有冇人識計通解呀????急

tan(30+x) tan(30-x) = 1-2cos2x
[(tan30+tanx)/(1-tan30tanx)][(tan30-tanx)/(1+tan30tanx)]= 1-2cos2x
[(1/√3+tanx)/(1-(1/√3)tanx)][(1/√3-tanx)/(1+(1/√3)tanx)]= 1-2cos2x
[(1+√3tanx)/(√3-tanx)][(1-√3tanx)/(√3+tanx)]= 1-2cos2x
(1-3tan^2x)/(3-tan^2x)= 1-2cos2x
(1-3tan^2x)=(3-tan^2x)( 1-2cos2x)
(1-3sin^2x/cos^2x)=(3-sin^2x/cos^2x)(1-4cos^2x+2)
(cos^2x-3sin^2x)=(3cos^2x-sin^2x)(3-4cos^2x)
(cos^2x-3sin^2x)=(3-3sin^2x-sin^2x)(4sin^2x-1)
(1-4sin^2x)=(3-4sin^2x)(4sin^2x-1)
(1-4sin^2x)=-16sin^4x+16sin^2x-3
16sin^4x-20sin^2x+4=0
4sin^4x-5sin^2x+1=0
(4sinx-1)(sinx-1)=0
sin^2x=1/4 or sin^2x=1
when sin^2x=1/4
sinx=1/2 or -1/2
x=180π+(-1)^n(30) or 180π+(-1)^n(-30)
when sin^2x=1
sinx=1 or -1
x=180π+(-1)^n(90) or 180π+(-1)^n(-90)

x=180π+(-1)^n(90)

tan(30°+x) tan(30°-x) = 1 - 2cos2x
[sin(30°+x) sin(30°-x)] / [cos(30°+x) cos(30°-x)] = 1 - 2cos2x [Use tan x = sin x / cos x]
[-1/2 {cos(2x) - cos(60°)} ] / [1/2{cos(60°) + cos(2x)] = 1 - 2cos2x
[cos(2x) - 1/2] / [cos(2x) + 1/2]= 1 - 2cos2x
(2y - 1)/(2y + 1) = 1 - 2y [Let y = cos 2x]
2(2y^2 - y - 1) = 0
(2y + 1)(y - 1) = 0
cos 2x = -1/2 or 1
2x = 360°n ± 120 or 360°n
x = 180°n ± 60 or 180°n (where n is any integer