{F4 A.Math} trigonometry

6 tan²A – 4 sin²A = 1
6 sin²A / cos²A - 4 sin²A = 1
6 sin²A - 4 sin²A cos²A = cos²A
6 sin²A - 4 sin²A (1 - sin²A) = 1 - sin²A
6 sin²A - 4 sin²A + 4 sin^4 A = 1 - sin²A
4 sin^4 A + 3 sin²A - 1 = 0
4 (sin²A)² + 3 sin² A - 1 = 0
(4 sin²A - 1)(sin²A + 1) = 0
4 sin²A = 1 or sin²A = -1 (Rejected)
sin²A = 1/4
sin A = 1/2 or sin A = -1/2
A = nπ + (-1)^n π/6 or A = nπ - (-1)^n π/6【If you want general solution】
A = π/6 or 5π/6 or A = 7π/6 or 11π/6【If you want 0≦A≦2π】


Hope it helps! ^^

2006-12-06 20:05:11 補充：
The key is to make tan, sec, csc to the basic form of sin and cos (tan A = sin A/cos A, sec A = 1/cosA, csc A = 1/sinA).Then use properties of sin and cos (sin²A cos²A = 1) to make it look like a polynomial and use the ways to solve a polynomial to solve the equation.

全條式乘 cos^2A，將所有cos^2A 代成 1-sin^2A，得出以下：

4sin^4A + 3sin^2A - 1 = 0

之後

(4 sin^2A - 1 ) (sin^2A + 1) = 0

後面個Bracket 是一個 Square 加一是零，不可能。
前面括號：

sinA = 1/2 or -1/2
So A = pi/6, 5pi/6, -pi/6, -5pi/6，之後每個都可以 + 2 pi n