Find the minimum positive period of the function y = 2Sin squarex + sin2x?

Find the minimum positive period of the function y = 2Sin squarex + sin2x?

y＝2sin^2x＋sin2x
y=-（1-2sin^2x)+sin2x+1
y=-cos2x+sin2x+1
y=V2sin(2x+3π/4）+1
So t = 2 π / 2 = π
So the minimum positive period is π

Let f (x) = 2sin2x + sin2x, X ∈ [0,2 π]. Find the set of X whose f (x) is positive

In the first method, one of the questions is: "f (x) = 1-cos2x + sin2x (2 points) = 1 + 2Sin (2x − π 4) (4 points) ∵ f (x) ⇔ 0 ⇔ 1 + 2Sin (2x ⇔ π 4) ⇔ 0 ⇔ sin (2x ⇔ π 4)

Given the function f (x) = 2Sin ^ 2x + sin2x (1), find the minimum positive period and maximum value of function f (x)

According to the multiple angle formula:
cos2x = 1-2sin²x
2sin²x = 1-cos2x
Qi
f(x)=1-cos2x+sin2x
=sin2x-cos2x+1
=√2[(√2/2)sin2x+(-√2/2)cos2x]+1
=√2sin[2x-(π/4)]+1
T = 2π/2 = π
Obviously, sin [2x - (π / 4)] ≤ 1
Therefore, the maximum value of F (x): √ 2 + 1

Why f (x) = sin2x-2sin? X = sin2x + (1-2sin? X) - 1 = sin2x + cos2x-1?

These are basic trigonometric functions, or identities
1-2sin^2x=cos2x

Vector M = (sin ω x + cos ω x, √ 3cos ω x), n = (COS ω x-sin ω x, 2Sin ω x) (ω > 0), the function f (x) = m * n + T, if f (x) is adjacent to each other on the image When x ∈ [0, π], the minimum value of function f (x) is 0 1. Find the expression of function f (x) 2. In the triangle ABC, if f (c) = 1 and 2sin2b = CoSb + cos (A-C), find the value of sina

f(x)=m×n+t
=cos²2ωx-sin²2ωx+2√3cosωxsinωx+t
=cos2ωx+√3sin2ωx+t
=2sin﹙30°＋2ωx﹚+t
The period is ω = 3 π t = 2
So f (x) = 2Sin (30 ° + 6 π x) + 2

(1 / 2) given the vector m (sin ω x, - √ 3cos ω x), n = (sin ω x, cos (ω x + π / 2)) (ω > 0), if f (x) = the minimum of M * n (1 / 2) given the vector m (sin ω x, - √ 3cos ω x), n = (sin ω x, cos (ω x + π / 2)) (ω > 0), if the minimum positive period of the function f (x) = m * n is

f(x)=m*n=sinωxsinωx-√3cosωx*cos(ωx+π/2)
=(1/2)(1-cos2ωx)+(√3/2)*2sinωxcosωx
=(√3/2)sin2ωx-(1/2)cos2ωx+1/2
=sin(2ωx-π/6)+1/2
If the minimum positive period T = 2 π / 2 ω = π
Then ω = 1

The function defined on R satisfies f (x + y) + F (X-Y) = 2F (x) f (y) f (0) ≠ 0, f (1 / 2) = 0. Let f (x) be a periodic function f (1 / 3) f (1 / 6) It is proved that f (x) is an even function and f (x) is a periodic function. If the function is monotone in [0, 1], we can find f (1 / 3) =? F (1 / 6) =? F (1 / 6) =? F (1 / 6) =? F (1 / 3) =? F (1 / 6) =? F (1 / 3) =?

Let x = 0, y = 0
f(0)+f(0)=2f(0)*f(0)
f(0)=1
Let x = 0
f(y)+f(-y)=2f(0)f(y)=2f(y)
F (- y) = f (y), so f is an even function
Let y = 1 / 2
f(x+1/2)+f(x-1/2)=2f(x)f(1/2)=0
-f（x+1/2）=f（x-1/2）=-f（x-3/2）
SO 2 is a cycle
Let x = 1 / 2, y = 1 / 2
f （1）+f(0)=2f(1/2)*f(1/2)=0
f(1)=-1
x=y=1/3
f（2/3）+f(0)=2f(1/3)f(1/3)
Let f (2 / 3) = m, f (1 / 3) = n
m+1=2mm
Let x = 2 / 3, y = 1 / 3
f（1）+f(1/3)=2f(2/3)f(1/3)
n-1=2mn
Find the equation
m、n
f（1/3）+f（0）=2f（1/6）f（1/6）
Thus, f (1 / 6) is obtained
Solve the equation yourself

It is known that the domain of F (x) is R. for any x, y ∈ R, there is f (x + y) + F (X-Y) = 2F (x) f (y), and f (0) ≠ 0. It is proved that y = f (x) is an even function

Let x = y = 0
Substitute f (x + y) + F (X-Y) = 2F (x) f (y)
F (0) + F (0) = 2F (0) f (0)
So f (0) = 1
f(x+y)+f(x-y)=2f(x)f(y)
Let x = 0
have to
f(y)+f(-y)=2f(0)f(y)=2f(y)
namely
f(y)=f(-y)
therefore
Y = f (x) is even function

Even function FX defined on R satisfies f (x + 1) = - f (x) period why is 2 Please make it clear, thank you

Let x = x 1, then the original formula can be changed to f [(x 1) 1] = - f (x 1) = f (x)

It is known that the function y = f (x) defined on R is even function, and x > = 0, f (x) = ln (x ^ 2-2x + 2), and the increasing interval of F (x)

Because: F (x) = LNX is a monotone increasing function in R domain
Where: let u (x) = x ^ 2-2x + 2
Then the original function f (x) = ln (U (x))
Therefore: the increasing interval of F (x) is the increasing interval of U (x)
How to find the increasing interval of U (x) = x ^ 2-2x + 2: draw the function graph of the function
The symmetric axis of U (x) function is x = 1, and the right half of the symmetry axis is the increasing region of the function
That is: [1, + infinity]
So the increasing interval of F (x) is [1, + infinity]