Do you know how tan x + 1 =sec x find the values of x超緊急(20)

1 + tanx = secx
( 1 + tanx )^2 = sec^2x
1 + tan^2x + tanx = sec^2x
sec^2x + tanx = sec^2x
tanx = 0
x = 0 degree, 180degree

by checking, x = 0 degree, 180 degree ( pi/2 )

2007-04-18 16:31:37 補充：
唔知你要general solution定係其他..

tan x + 1 = sec x
(sin x / cos x) + 1 = (1 / cos x)
sin x + cos x = 1　＜＝[左右乘cos x,cos x≠0]
[註 : 因為tan x同sec x都以cos x為分母 , 所以cos x≠0]
sqrt(2) [(1/sqrt(2) sin x + 1/sqrt(2) cos x] = 1 ＜＝[subsidiary angle form,輔助角]
sin 45° sin x + cos 45° cos x = 1/sqrt(2)
cos (x - 45°) = cos 45°
x - 45° = 360°n±45°
x = 360°n+90°(rej.) or x = 360°n　＜＝[cos x≠0 => x≠90°]
Overall, x = 360°n , n=1,2,3...