The equation of symmetry axis of image with function y = 3sin (2x + π / 4)

The equation of symmetry axis of image with function y = 3sin (2x + π / 4)

2x+π/4=kπ+π/2
X = k π / 2 + π / 8 (k is an integer)

Find the axis of symmetry equation, center of symmetry and monotone increasing interval of function y = 3sin (2x + π / 4) - 2

The axis of symmetry of SiNx is x = π / 2 + K π
The center of symmetry is (k, π, 0)
Monotone increasing interval is (- π / 2 + 2K π, π / 2 + 2K π)
For the function y = 3sin (2x + π / 4) - 2, just substitute (2x + π / 4) into the position of X and find out the range of X. note that the ordinate of the center of symmetry is - 2

It is known that the domain of F (x) = asin (2x + Pai / 3) + 1 (a > 0) is r, and if - 7pai / 12

Let 2x + π / 3 = π / 2 + K π, K ∈ Z, then x = (6K + 1) π / 12,
Because x ∈ [- 7 π / 12, - π / 12], k = - 1,2x + π / 3 = - π / 2
a=-1

It is known that the period of the function f (x) = asin (ω x + φ), X ∈ R (where a > 0, ω > 0, 0 < φ < π / 2) is π and the lowest point m (π / 12) on the image is π
The set of values of X when finding the minimum value of F (x)
Finding monotone interval of F (x)

The period is π, so ω = 2
The set of minimum values is {π / 12 + K π, K ∈ n +}
Monotone decreasing interval (- 5 π / 12 + K π, π / 12 + K π)
Monotone increasing interval (π / 12 + K π, 7 π / 12 + K π) (K ∈ n +)