The period of the function y = 1 + cos4x, X ∈ R is______ The even function of

The period of the function y = 1 + cos4x, X ∈ R is______ The even function of

Function y = 1 + cos 4x, X ∈ R, even function with period π / 2
(2π)/4=π/2

1 "y = SIN3 / 4x R 2" y = cos4x R 3 "y = 1 / 2cosx R 4"
To find the period of the following function, write the procedure (do not set). 1 "y = SIN3 / 4x R
2"y=cos4x R
3 "y = 1 / 2cosx R 4" y = sin (1 / 3x + Wu / 4) r

1. Period T = 2 π / (3 / 4) = 8 π / 3
2. Period T = 2 π / 4 = π / 2
3. Period T = 2 π / 1 = 2 π
4. Period T = 2 π / (1 / 3) = 6 π

If f (x) = sin3x + | sin3x | is a periodic function, what is the minimum positive period?
Why
Why 2 π / 3?

When sin3x < 0, FX = 0, when sin3x ≥ 0, FX = 2sin3x, then FX is actually a piecewise function, write it out, and then look at when sin3x < 0 (with a period T1) and when sin3x ≥ 0 (with a period T2). In fact, it can be seen, because Sint itself is a period function (can be divided into positive and negative parts), so the piecewise function after participation still has a period, And the period T = T1 + T2
0 2kπ/3~~~~~(2k+1)π/3 T1=π/3
f（x）=
2sin3x (2k+1)π/3~~~~~(2k+2)π/3 T2=π/3
So t = 2 π / 3

Find the period of the following function, 1. Y = sin3x, X belongs to R2. Y = 3sinx / 4, X belongs to R3. Y = 2Sin (2x - π / 6)

1. Y = sin3x can be understood as y = SiNx's ordinate unchanged, abscissa changed to the original 1 / 3, so the period also changed to the original 1 / 3, that is 2 π / 3.2. Y = 3sinx / 4 can be understood as y = SiNx's ordinate extended to the original 3 times, abscissa changed to the original 4 times, so the period also changed to the original 4 times, that is 8 π. 3. Y = SiNx

Period of function y = sin3x

The period of trigonometric function T = 2 π / W
W = 3 / 4 in y = sin3x
T=2π/3/4
T=8π/3

(1)y=sin2/3x,x∈R(2)y=1/2cos4X,X∈R
Detailed process!

Are you asking for a cycle
(1) The period of SiNx is 2 π
So sin2 / 3x period is 2 π / (2 / 3)
=3π
(2) The period of cosx is 2 π
Y = 1 / 2cos4x period
=2π/4
=π/2

Find the period of the following functions: (1) y = sin23x, X ∈ R; (2) y = 12cos4x, X ∈ R

（1）∵y=sin23x，∴T=2π23=3π；（2）∵y=12cos4x，∴T=2π4=π2．

Find the period of the following functions: (1) y = sin23x, X ∈ R; (2) y = 12cos4x, X ∈ R

（1）∵y=sin23x，∴T=2π23=3π；（2）∵y=12cos4x，∴T=2π4=π2．

The least positive period of the function y = cosx - SiNx is (seeking process)

y=cos²x－sin²x
=cos(2x)
The minimum positive period is 2 π / 2 = π

If the minimum positive period of the function y = sinwx times coswx is 4 π, then the value of the positive number W is

The result of y = sinwx multiplied by coswx is 0.5sin2wx, so the minimum positive period is t = 2 π divided by 2W = 4 π, and the solution is w = 1 / 4,