F (x) is an even function defined on R when x

F (x) is an even function defined on R when x

Let g (x) = XF (x)
Then we get the following results:
g'(x)=f(x)+xf'(x)
g(-4)=-4f(-4)=0
G (x) is an odd function
So the title becomes:
G (x) is an odd function

It is known that even function f (x) defined on R is a decreasing function on (- ∞, 0]. If f (1 / 2) = 0, then the solution set of inequality f (log4x) > 0 is

I write in detail, you give me the best
If the even function f (x) defined on R is a decreasing function on (- ∞, 0], then f (x) increases on (0, + ∞)
(log4x) = k, then if f (k) > 0, then
k(1/2)
When k

It is known that even function f (x) defined on R is a decreasing function on (negative infinity, 0]. If f (1 / 2) = 0, find the solution set of inequality f (log4x) > 0

It is known that even function f (x) defined on R is a minus function on (negative infinity, 0]
Then f (x) is an increasing function at [0, + ∞)
It is known that f (1 / 2) = 0
Then the inequality f (log4x) > 0 = f (1 / 2)
According to the definition of increasing function, log4 x > 1 / 2
So x > 4 ^ (1 / 2)
X > 2

Sincox + 3 * 3?

5 * cosx = 3 (√ 15 * SiNx + 3 √ 5 * cosx = 3 (√ 15 * SiNx + 5 * cosx = 3 (√ 15 * SiNx + 5 * cosx = 3 (√ 15 * SiNx + 5 * cosx = 3 (√ 5 * SiNx + 1 / 2cosx) = 6 √ 5 (COS π / 6 * SiNx + SiNx + 1 / 2cosx) = 6 √ 5 (COS π / 6 * SiNx + SiNx + SiNx + 1 / 2cosx) = 6 √ 5 (COS π / 6 * SiNx + SiNx + SiNx + SiNx + 1 / 2cosx) = 6 √ 5 (COS π / 6 * SiNx + 6 * SiNx + SiNx + SiNx + SiNx + 1 / 2cosx) = 6 √ 5 π / 6cosx) = 6 √ 5sin (x + π / 6)

How to convert the sum of sin and COS into asin (ω x + φ)?

tan(φ) = B/A φ = arctan(B/A)
sin(φ) = B/√(A² + B²)
cos(φ) = A/√(A² + B²)
The angles of asin (ω x) + bcos (ω x), sine and cosine are the same
= √(A² + B²)[A/√(A² + B²) * sin(ωx) + B/√(A² + B²) * cos(ωx)]
= √(A² + B²)[cos(φ)sin(ωx) + sin(φ)cos(ωx)]
=It can also be expressed as
= √(A² + B²)sin[ωx + arctan(B/A)]

How to transform the function y = 2sinxcosx + cos? X-sin? X into the form of y = asin (ω x + φ)?

Let me answer
According to the formula of double angle,
sin2α = 2cosαsinα
cos2α = (cosα)^2 − (sinα)^2
therefore
y=sin2x+cos2x
y=√2sin(2x+π/4)

Y = cos ^ 4x + sin ^ 4x-3 is reduced to y = asin (Wx + F)

y=(sin^2x+cos^2x)^2-3-2sin^2xcos^2x=(cos4x-9)/4

It is known that the function y = sin ^ 2 x + 2 SiN x cosx + 3 cos ^ 2 x, X ∈ R The function y = sin ^ 2 x + 2 SiN x cosx + 3 cos ^ 2 x, X ∈ R (1) is the minimum positive period of a function? (2) in what interval is a function increasing? (3) how can the image of the function be transformed from the image of y = √ 2sin2x, X ∈ r?

The key is the first step,
y＝sin^2 x＋2sinxcosx＋3cos^2 x=1+2cos^2 x+2sinxcosx=cos2x+sin2x=√2sin（2x+π/4）
1）T=2π/w=π
2) - π / 2 + 2K π ≤ 2x + π / 4 ≤ π / 2 + 2K π, K ∈ Z
3) This is also a fallible point, y = √ 2Sin (2x + π / 4) = = √ 2sin2 (x + π / 8), so we should draw two
Therefore, the image of the function can be obtained by the left shift π / 8 of y = √ 2sin2x, X ∈ R

Y = radical 2Sin (2x - π) cos [2 (x + π)] is reduced to asin (Wx + y)

sin(2x-π)=-sin(2x)
cos【2(x+π)】=cos(2x)
y=√2sin(2x-π)cos【2(x+π)】
=-√2sin(2x)cos(2x)
=-√2/2 * sin(4x)
=√2/2 * sin(4x+π)

Proof of sin? α + cos? α = 1

According to the definition, take a point P (x, y) on the unit garden
sinA=y/r,cosA=x/r
(sinA)^2=y^2/r^2,(cosA)^2=x^2/r^2
(sinA)^2+(cosA)^2=y^2/r^2+x^2/r^2
=(y^2+x^2)/r^2
Because R ^ 2 = y ^ 2 + x ^ 2
So (Sina) ^ 2 + (COSA) ^ 2 = 1