Let f (x) = SiNx + sin (x + π 3). Find the minimum value of F (x) and the set of X that makes f (x) the minimum value

Let f (x) = SiNx + sin (x + π 3). Find the minimum value of F (x) and the set of X that makes f (x) the minimum value

F (x) = SiNx + sin (x + π 3) = SiNx + sinxcos π 3 + cosxsin π 3 = 32sinx + 32cosx = 3sin (x + π 6) ≠ when x + π 6 = 3 π 2 + 2K π (K ∈ z), that is, x = 4 π 3 + 2K π (K ∈ z), the minimum value of F (x) is - 3, so the minimum value of F (x) is - 3, then the set of X {x | x = 4 π 3 + 2K

How can y = (1 / 2) sin (2x + π / 6) + 1 be obtained by y = SiNx (x ∈ R) transformation?

The ordinate does not change, and the left translation is π / 6 units
And then x is reduced to twice the original
The abscissa does not change and the ordinate reduces to the original two
Then the whole function moves up one unit

How can y = sin (2x + π / 6) + 1 / 2 be obtained by y = SiNx (x ∈ R) transformation?

Method 1 S1 moves π / 6 to the left / / changes to y = sin (x + π / 6) S2, the abscissa of all points changes to the original 1 / 2, and the ordinate remains unchanged / / changes to y = sin (2x + π / 6) S3, moves up 1 / 2 / / changes to y = sin (2x + π / 6) + 1 / 2. Method 2 (upstairs wrong, help him to correct it) reduces the SiNx image twice to sin (2x) and then moves to the left

Graph (1) y = 1-sinx (2) y = sin (- x)
Hurry! Functions don't work

This is easy to do. First, analyze the range of SiNx, and then transform it according to the requirements of the topic. First, calculate the range of SiNx from - 2 π to 2 π, which is [- 1,1], then subtract the range of SiNx from 1, and the calculation result is still [- 1,1]. This is the calculation result of the first question, and then when drawing, first