2Sin (2 * π / 3 + θ) + 2 = 4, finally we get π / - 6 + 2K π. How can we get it! Given that the maximum value of the function y = sin (W + θ) + m is 4, the minimum value is 0, the minimum positive period is π / 2, and π / 3 is an axis of symmetry of its image, then the analytic expressions satisfying the conditions in the following expressions are

2Sin (2 * π / 3 + θ) + 2 = 4, finally we get π / - 6 + 2K π. How can we get it! Given that the maximum value of the function y = sin (W + θ) + m is 4, the minimum value is 0, the minimum positive period is π / 2, and π / 3 is an axis of symmetry of its image, then the analytic expressions satisfying the conditions in the following expressions are

Sin (2 π / 3 + x) = 12 π / 3 + x = 2K π + π / 2 Result x = 2K π - π / 6
I didn't give you all the back one

Given sin (α - 3 π) + cos (π / 2 + α) = m, find the value of COS (α - 7 π / 2) + 2Sin (2k π - α)

sin（α-3π）+cos(π/2+α)=m,
-sinα-sinα=m
So: sin α = - M / 2
cos（α-7π/2）+2sin（2kπ-α）
=-sinα-2sinα
=-3sinα
=3m/2

When α ∈ {2K π + π / 4,2k π + 5 π / 4} (K ∈ z), sin α - cos α ≥ 0,
The original formula = √ 2-m ^ 2!

〔sinα－cosα）^2
=1-2sinαcosα
=1-sin2α
If α ∈ {2K π + π / 4,2k π + 5 π / 4} (K ∈ z), then 2 α ∈ R
1-sin2α∈[0,2]
〔sinα－cosα）^2=1-sin2α=√2-m^2∈[0,2]
And then it's easy. Let's do it by ourselves