Given function f (x) = sin (2x + φ) + ACOS (2x + φ) Given the function f (x) = sin (2x + α) + ACOS (2x + α), where a > 0 and 0 < a < π, if the image of F (x) is symmetric with respect to the straight line x = π / 6, and the maximum value of F (x) is 2. (1) finding the values of a and α (2) how to get the image of y = 2Sin (2x + π / 3) from the image of y = f (x)

Given function f (x) = sin (2x + φ) + ACOS (2x + φ) Given the function f (x) = sin (2x + α) + ACOS (2x + α), where a > 0 and 0 < a < π, if the image of F (x) is symmetric with respect to the straight line x = π / 6, and the maximum value of F (x) is 2. (1) finding the values of a and α (2) how to get the image of y = 2Sin (2x + π / 3) from the image of y = f (x)

(1) 1 * 1 + A * a = 2 * 2 is obtained from the maximum value of 2, and all a values are root sign 3. 2 * π / 6 + α + π / 3 = (n + 1 / 2) π is obtained by simplification, and α = 5 π / 6 is obtained according to the value range. (2) firstly, the period of the function is shortened to half of the original period, and then the length of π / 6 is shifted to the left, and the ordinate is extended twice

Given function f (x) = sin (2x + π / 3)
(1) After the image of function f (x) is shifted a (a > 0) units to the right, the image of function g (x) is obtained, and the function g (x) is an even function
(2) Find the sum of the roots of the equation f (x) = 0 in (0, π)
(3) The equation f (x) + 2m = 0 has exactly two on [0, π / 2], so we can find the value range of real number M

Given the function g (x) = 1 / 2Sin (2x + 2 π / 3) f (x) = ACOS ^ 2 (x + π / 3) + B, and the image of function y = f (x) is the image of function y = g (x)
According to the translation of vector a = (- π / 4,1 / 4). (1) find the values of real numbers a and B. (2) let H (x) = g (x) - root 3f (x), find the minimum value of H (x) and the corresponding value of X

Because [x + √ (X & # 178; + 1)] [- x + √ (X & # 178; + 1)] = 1, that is: - x + √ (X & # 178; + 1) = 1 / [x + √ (X & # 178; + 1)],
Therefore, LG [- x + √ (X & # 178; + 1)] = - LG [x + √ (X & # 178; + 1)];
Let g (x) = f (x) - 2 = x & # 179; + LG [x + √ (X & # 178; + 1)],
Then G (- x) = (- x) &# 179; + LG {- x + √ [(- x) &# 178; + 1)]} = - X & # 179; + LG [- x + √ (X & # 178; + 1)] = - X & # 179; - LG [x + √ (X & # 178; + 1)] = - G (x), = -,
It is known that f (x) has a minimum value of - 5 on (- ∞, 0),
The results show that G (x) = f (x) - 2 has a minimum value of - 5-2 = - 7 on (- ∞, 0);
Because g (x) is an odd function,
So, G (x) has a maximum value of 7 on (0, + ∞),
It can be concluded that f (x) = g (x) + 2 has a maximum value of 7 + 2 = 9 on (0, + ∞)