一道求值域和定义域

根号下大于等于0
2sinx+1>=0
sinx>=-1/2
sin(2kπ-π/6)=sin(2kπ+7π/6)=-1/2
所以2kπ-π/6<=x<=2kπ+7π/6
定义域[2kπ-π/6,2kπ+7π/6]

1/9<=x<=9
3^(-2)<=x<=3^2
log3(x)底数大于1，是增函数
所以log3[3^(-2)]<=log3(x)<=log3(3^2)
-2<=t<=2

1.2sinx+1≥0
sinx≥-1/2
x∈（2kπ，5π/6+2kπ），k是整数
2.当x=1/9，log3x=log3（1/9）=-2
当x=9，log3x=log39=2
∴设log3x=t后，x的定义域变成[-2,2]

1，
根号下大于>=0
即2sinx+1>=0
所以sinx>=-1/2
所以2kπ-π/6<=x<=2kπ+7π/6
所以定义域是[2kπ-π/6,2kπ+7π/6]

2，
1/9<=x<=9
log3(x)底数大于1，是增函数
log3(1/9)<=log3(x)<=log3(9)
即log3(1/9)<=t<=log3(9)
即-2<=t<=2