2、解不等式绝对值arcsin(x-1)>π/63、sinx+cosx=0在[0,π]上有相等2实根,则x的取值范围如何?

Question

2、解不等式
绝对值arcsin(x-1)>π/6
3、sinx+cosx=0在[0,π]上有相等2实根,则x的取值范围如何?

Answer 1

arcsin(x-1)>π/6
x-1>sin(π/6)=1/2
x>3/2
sinx+cosx=0
sinx=-cosx
sinx=cos(π-x)=sin(x-π/2)
因为x属于[0,π]
所以x+(x-π/2)=π,解得x=3π/4