引用:
原帖由 andydison 於 2013-6-22 11:10 AM 發表
已知 \( 0 \leq x \leq\frac{\pi}{2}\) 且 \(\sin^8 x + \cos^8 x = \frac{97}{128} \),求 x 的兩個解
令 (sinxcosx)^2 = t
0 ≦ x ≦π/2,0 ≦ t ≦ 1/4
(sinx)^8 + (cosx)^8 = 2t^2 - 4t + 1 = 97/128
t = 1/16
sinxcosx = 1/4
sin2x = 1/2
x = (1/12)π or (5/12)π