The range of function y = sin2x + 2sinx + 1

The range of function y = sin2x + 2sinx + 1

When the derivative 2cos2x + 2cosx = 0,
cos2x=cosx
cosx*cosx-sinx*sinx=cosx
Add 1 to both sides to get
2cosx*cosx=cosx+1;
The following steps are relatively simple, that is to find the solution of the equation and take it to y. I hate calculation, and it's not convenient for computer to type

Find the definition field of sin2x under y = root
The maximum and minimum of y = 2cos ^ 2x + 5sinx-4
The range of cosx 1 / 2 x

1. From sin2x ≥ 0
x∈[kπ,π/2+kπ]
2.y=2(1-sin^2x)+5sinx-4
=-2(sinx-5/4)^2+9/8
Because - 1 ≤ SiNx ≤ 1, y ∈ [- 45 / 4,5 / 4]
3. It's not very clear. You can draw the graph of two functions and then get it

The definition field of function f (x) = 1 / root sign sin2x is

Denominator not zero √ sin2x ≠ 0
Root sign sin2x ≥ 0
That is sin2x > 0
The solution is x ∈ (K π, K π + π / 2) k ∈ Z

Given that the m power of 5 is 4 and the n power of 5 is 3, find the value of 2m + 3N power of 5

2m + 3N power of 5
=2m power of 5 × 3N power of 5
=(m power of 5) square × (n power of 5) cube
=16×27
=432