If x∈[0, π/2], then 0≤sinx≤ 1,
Let sinx=t, t∈[0, 1],
Function y=-(t-p)? +p？ +q+ 1 is a quadratic function about the downward opening of t, and the symmetry axis is t=p(t∈[0, 1]).
Suppose there are p∈Z and q∈Z so that the function y=-(t-p)? +p？ The value range of +q+ 1(t∈[0, 1]) is [7, 10].
When p≥ 1 and p∈Z, y=-(t-p)? +p？ +q+ 1 is a monotonically increasing function on [0, 1].
What is the function value when the maximum value is t= 1 =- 1 +2p× 1+q+ 1 = 2p+q = 10，
What is the function value when the minimum value is t=0 =-0? +2p×0+q+ 1=q+ 1=7，
The solution is q = 6, p = 2 >；; 1,
When p≤0 and p∈Z, y=-(t-p)? +p？ +q+ 1 is a monotonically decreasing function on [0, 1].
What is the function value when the maximum value is t=0 =-0? +2p×0+q+ 1 = q+ 1 = 10，
What is the function value when the minimum value is t= 1 =- 1 +2p× 1+q+ 1=2p+q=7，
Q = 9，p =- 1
To sum up, there are p=2, q=6, or p=- 1, q=9, which makes the range of f(x) [7, 10].