Find the monotone increasing interval and monotone decreasing interval of sin (2x + π - 4) at the bottom of function y = log2

Find the monotone increasing interval and monotone decreasing interval of sin (2x + π - 4) at the bottom of function y = log2

Y = log2 and sin (2x + π / 4) are significant
sin(2x+π/4)＞0
The single increasing interval is
2kπ＜2x+π/4＜2kπ+π/2
2kπ-π/4＜2x＜2kπ+π/2-π/4
kπ-π/8＜x＜kπ+π/8
The single minus interval is
2kπ+π/2＜2x+π/4＜2kπ+π
2kπ+π/2-π/4＜2x＜2kπ+π-π/4
kπ+π/4＜x＜kπ+3π/8

The monotone decreasing interval of the function y = log2 sin (2x + 6 / Π) is A〔k∏-∏/12,k∏+5∏/12〕(k∈Z) B〔k∏+∏/6,k∏+2∏/3〕(k∈Z) C〔k∏-∏/3,k∏+∏/6〕(k∈Z) D〔k∏+∏/6,k∏+5∏/12〕(k∈Z)

The monotone decreasing interval of DSIn (2x + 6 / Π) is (K Π + Π / 6, K Π + 2 Π / 3), sin (2x + 6 / Π) / 6, K Π + 5 Π / 12), sin (2x + 6 / Π) is negative at (k Π + 5 Π / 12, K Π + 2 Π / 3), sin (2x + 6 / Π) is negative

Find the monotone increasing interval of function y = log2 sin (x + π / 3)

The domain must be sin (x + π / 3) > 0, that is 2K π

What is the value range of the function y = sin ^ 4x + cos ^ 4x? Simplify the function first

y=sin^4x+cos^4x=[(sinx)^2+(cosx)^2]^2-2(sinxcosx)^2
=1-2(sinxcosx)^2
=1-[(sin2x)^2]/2
=1-[1-(cos4x)]/4
=(3+cos4x)/4
Max = 1
Minimum = 1 / 2
Range [1 / 2,1]

The value range of the function cos ^ 4x sin ^ 4x is_____ The period is_____ .

cos^4x-sin^4x
=(cos^2x+sin^2x)(cos^2x-sin^2x)
=cos2x
So: the range is [- 1,1] and the period is: T = k π
The minimum positive period is π

Find the function f (x) = sin ^ 4x + cos ^ 4x + sin ^ 2xcos ^ 2x / 2-2sinxcosx-1 / 2sinxcosx + 1 / 4cos ^ 2x Find the minimum positive period, maximum value and minimum value of the function f (x) = (sin ^ 4x + cos ^ 4x + sin ^ 2xcos ^ 2x) / (2-2sinxcosx) - (1 / 2sinxcosx) + (1 / 4cos ^ 2x)!

Simplify first
f(x)=（sin^4x+cos^4x+sin^2xcos^2x）/（2-2sinxcosx）-1/2sinxcosx+1/4cos^2x
=【（sin²x+cos²x）²-sin²xcos²x】/（2-sin2x）-1/4*sin2x+1/4*（1+cos2x）/2
=（1-1/4*sin²2x）/（2-sin2x）-1/4*sin2x+1/8+（cos2x）/8
=1/4（4-sin²2x）（2-sin2x）-1/4*sin2x+1/8+（cos2x）/8
=1/4（2+sin2x）-1/4*sin2x+1/8+（cos2x）/8
=（5+cos2x）/8.
Then the minimum positive period of F (x) is π and the maximum value is (5 + 1) / 8 = 3 / 4,
The minimum value is (5-1) / 8 = 1 / 2

Find the minimum positive period, maximum value and minimum value of the function FX = sin ^ 4x + cos ^ 4x + sin ^ 2xcos ^ 2x / 2-2sinxcosx

F (x) = sin ^ 4x + cos ^ 4x + sin ^ 2xcos ^ 2x / 2-2sinxcosx = (sin? 2x + cos? X)? - 3sin? Xcos? X / 2-2sinxcosx = 1-3sin? 2x / 8-sin2x = - (3 / 8) (sin2x + 4 / 3)? + 5 / 3. Obviously, the minimum positive period is 2 π / 2 = π

Find the function y = sin ^ 4x + cos ^ 4x + sin ^ 2x cos ^ 2x divided by the minimum positive period of 2-sin2x, the maximum value and the minimum value

Y=（sin^4x +cos^4x +sin^2x cos^2x） /（2-sin2x） =（(sin^2x+cos^2x)^2-sin^2x cos^2x)/（2-sin2x） =(1-sinxcosx)(1+sinxcosx)/2(1-sinxcosx) =(1+sinxcosx)/2 = 1/2+sin2x/4sin2x=0 Ymin=1/2sin2x=1 ...

Find the minimum positive period and range of sin ^ 4x + cos ^ 4x + 4sin ^ 2xcos ^ 2x - 1

y=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1
=1+2sin^2xcos^2x-1
=2sin^2xcos^2x
=sin^2(2x)/2
=(1-cos4x)/4
The period is obviously pi / 2, not pi / 4
The range is: [0,1 / 2]

Find the minimum positive period and range of sin ^ 4x + cos ^ 4x + 4sin ^ 2xcos ^ 2x - 1

2X ^ 2x-1 = (sin ^ 2x + cos ^ 2x ^ 2x-1 = (sin ^ 2x + cos ^ 2x ^ 2 ^ 2 + 2Sin ^ 2xcos ^ 2x-1 = 1 + 2Sin ^ 2xcos ^ 2x-1 = 2Sin ^ 2xcos ^ 2x = 1 / 2 * sin ^ 2 (2x) = (1-cos4x) / 4 = 1 / 4-1 / 4 * cos4x cycle T = 2 π / 4 = π / 2 value domain is: [0,1 / 2] if you agree with my answer, please click the left left left left left click if you agree with my answer, please click the left left left left, if you agree with my answer, please click the left left left left left, if you agree with my answer, please click the left left left left left click if you agree with my answer, please click left left lower corner