三角函数 问题

無聊人

我想看看大家怎么解这题，我是用2x为重点。

Solve the following equation, 2sin-1 (x) + sin-1 (2x) = pi/2.

Revenge38

trigoometry rite???

but i dont understand the question ....

can u explain??

無聊人

sin-1 x 并非 cosec x

sin-1 x = theta (angle来的)
x 则是比例。

pi/2 --> radian mode

fadeev_popov

2arcsin(x) = pi/2 - arcsin(2x)
cos[2arcsin(x)] = cos[pi/2 - arcsin(2x)]
cos^2[arcsin(x)] - sin^2[arcsin(x)] = sin[arcsin(2x)]
(1 - x^2) - x^2 = 2x
2x^2 + 2x - 1 = 0
x = (1/2)(-1 + sqrt(3))