Monotone decreasing interval of function f (x) = sin (x + Pai / 6) + sin (x-pai / 6) + cosx minimum positive period sum on [0,2pai]

F (x) = SiNx * radical 3 / 2 + 1 / 2cosx + SiNx * radical 3 / 2-1 / 2cosx + cosx
=Root sign 3sinx + cosx
=2sin(x+Pai/6)
So the minimum positive period T = 2pai / 1 = 2pai
0

Given function f (x) = sin (3pai / 2-x) cosx sinxcos (PAI + x)
(1) (2) if a is an acute angle, f (a) = 0, BC = 2, find the length of AC

F (x) = sin (3pai / 2-x) cosx sinxcos (PAI + x) = - cosx * cosx SiNx * (- cosx) = - (1 + cos2x) / 2 + (1 / 2) sin2x = (1 / 2) sin2x - (1 / 2) cos2x-1 / 2 = (√ 2 / 2) * [sin2x * cos (π / 4) - cos2x * sin (π / 4)] - 1 / 2 = (√ 2 / 2) sin (2x - π / 4) - 1 / 2 (1) increasing interval 2K - π /

Find the function f (x) = cosx * 2 + 2asinx-1 (0 ≤ x

f(x)=cosx*2+2asinx-1=1-sinx^2+2asinx-1=-sinx^2+2asinx
Because 0 ≤ X

RELATED INFORMATIONS

1. The known function f (x) = 2asinx * cosx + 2cos ^ 2x + 1, f (π / 6) = 4 Find the real number a Finding the symmetric center coordinates of the function f (x) image Finding the range of function f (x) on [- π / 4, π / 4]
2. The piecewise function f (x) = {x2-x + 1 (x > = 2) x + 1 (x) is known
3. Draw the program block diagram of the function f (x) = x + 1
4. Let the maximum value of the function y = x ^ 3-2x + A in the interval [1,2] be 8, then what is the value of a
5. The quadratic function y = - x squared - 4x-5 is known (1) Translate the image of the quadratic function up and down so that its vertex just falls on the image of the positive scale function y = - x, and find the analytic expression of the quadratic function at this time
6. The quadratic function y = 2x ^ - 8x + 3 is transformed into y = a (x + H) ^ 2 + K by collocation method
7. (1) the quadratic function y = x2-4x + 3 is changed into the form of y = (X-H) 2 + K by the collocation method;
8. Given the function f (x) = x ^ 2-2x + 2, y = f (x), the minimum value on [T, t + 1] is the function g (x) of T, the analytic expression of G (x) is obtained
9. Finding the minimum value of function f (x) = LG (x ^ - 2x + 2)
10. Given the function f (x) = 2 + log3x (1 ≤ x ≤ 9), find the maximum and minimum value of the function y = [f (x) ^ 2 + F (x ^ 2), and find the corresponding value
11. Let f (x) (x ∈ R) be a periodic function with Z as the minimum positive period, and f (x) = (x-1) ^ 2 when x ∈ [0,2]. Find the values of F (3), f (7 / 2),
12. Which of the following functions are odd or even? Which are not odd or even? (1) f (x) = 5x + 3 (2) f (x) = 5x (3) f (x) = x & # 178; + 1
13. Let even function f (x) satisfy f (x) = x ^ 3-8 (x > = 0), if f (X-2) > 0, then the value range of X is? Did you answer that before? I can't see it. Can you answer it again
14. Let even function f (x) satisfy f (x) = 2 & # 710; x-4 (x ≥ 0), then {x| f (X-2) > 0}= 1. {x | x 〈 - 2 or X 〉 4} 2. {x | x ＜ 0 or X ＞ 4} 3. {x | x ＜ 0 or X ＞ 6} 4. {x | x ＜ - 2 or X ＞ 2}
15. Let f (x) and G (x) be the even and odd functions of R respectively in the domain of definition, f (x) = 2F (x) + G (x) = MX ^ 2 + NX + 1, f (1) = 1, f (- 1) = 5, M, n Finding f (2) and G (2)
16. If the domain of F (X & # 178;) is [- 1,1], the domain of F (x) is
17. If the definition field of function f (X & # 178; - 1) is [- 1,2], then the value range of X in function f (√ x) is X ∈ [- 1,2] → X & # 178; - 1 ∈ [- 1,3], ■ √ x ∈ [- 1,3] → x ∈ [0,9]. I don't understand how X & # 178; - 1 ∈ [- 1,3], and then get x ∈ [0,9] It's a bit confusing. Is the analytic expression of F (X & # 178; - 1) and f (√ x) the same, with the same range, but the whole (X & # 178; - 1, √ x) as X is different, so the definition field of X in that whole is required
18. Given the function f (x) = 2 + log3x (1 ≤ x ≤ 9), find the maximum and minimum value of the function y = [f (x) ^ 2 + F (x ^ 2), and find the corresponding value Can you add the value of F (x ^ 2) to the value of y = f (x) ^ 2
19. Let f (x) be an even function of the domain of definition on R, and if x > 0, f (x) = x (X-2), then find if X
20. It is known that the function f (x) whose domain is R is an even function. When x ≥ 0, f (x) = (x + 1) divided by X, it is proved that f (x) = 2 ^ (1-x) has a solution in the interval (1,2)

Monotone decreasing interval of function f (x) = sin (x + Pai / 6) + sin (x-pai / 6) + cosx minimum positive period sum on [0,2pai]