sin눀a+(3sina-根号10)눀=1?

2025-02-27 00:18:00
推荐回答(2个)
回答1:

令t=sina,则-1<=t<=1
t^2+(3t-√10)^2=1
t^2+9t^2-6√10t+10=1
10t^2-6√10t+9=0
(√10t-3)^2=0
t=3/√10
sina=3/√10
a=2kπ+arcsin(3/√10)或(2k+1)π-arcsin(3/√10),其中k是任意整数

回答2:

sin²a+(3sina-√10)²=1
sin²a+9sin²a-6√10sina+10=1
10sin²a-6√10sina+9=0
(√10sina-3)²=0
√10sina-3=0
sina=3/√10=3√10/10