中四M2數學 急!

(csc x)^2 = 25(sin x)^2 - 20(cot x)^2
(sin x)^2(csc x)^2 = (sin x)^2[25(sin x)^2 - 20(cot x)^2]
1 = 25(sin x)^4 - 20(cos x)^2
25(sin x)^4 - 20[1 - (sin x)^2] - 1 = 0
25(sin x)^4 + 20(sin x)^2 - 21 = 0
[5(sin x)^2 + 7][5(sin x)^2 - 3] = 0
(sin x)^2 = -7/5 (rejected) or (sin x)^2 = 3/5
sin x = +/- sqrt(3/5)

Answers are depends on the range of x

csc^2x=25sin^2x-20cot^2x

> 1 = 25(sinx)^4 - 20(cosx)^2

> 1 = 25(sinx)^4 -20[1-(sinx)^2)]

> 25(sin^2x)^2 + 20sin^2 - 21 = 0

> (sinx)^2 = -7/5 (rej) or (sinx)^2 = 3/5

> sinx = +-sqroot(3/5)

csc^2 x = 25sin^2 x - 20cot^2 x
1 = 25sin^4 x - 20cos^2 x
25sin^4 x - 20 + 20sin^2 x - 1 = 0
25sin^4 x + 20sin^2 x - 21 = 0
Sub sin^2 x = y, then the question will be solved

20cot^x

??!!!!