Given,
f(x)=sin−1(x2+2x+7x2−3x+2)
We know that domain of sin−1(f(x)) is −1≤f(x)≤1
So, x2+2x+7x2−3x+2≥−1 and x2+2x+7x2−3x+2≤1
Now solving x2+2x+7x2−3x+2≥−1
Since denominator x2+2x+7 is always positive,
So, x2−3x+2≥−(x2+2x+7)
⇒2x2−x+9≥0
⇒x∈R.....(1)
Now solving x2+2x+7x2−3x+2≤1
⇒x2−3x+2≤x2+2x+7
⇒5x≥−5
⇒x≥−1.....(2)
So from equation (1)&(2) we get domain x∈[−1,∞)