The period of the function f (x) = sinxcosx + cos ^ 2x is

The period of the function f (x) = sinxcosx + cos ^ 2x is

f(x)=1/2 sin2x+1/2 cos2x+1/2
=1/2（sin2x+cos2x）+1/2
According to the auxiliary angle formula: F (x) = radical 2 / 2Sin (2x + π / 4) + 1 / 2
So the minimum positive period is t = 2 π / 2 = π

Known function f (x) = sinxcosx + cos ^ 2x 1: finding the minimum period of a function X2: Find the maximum value of the function and the set of the corresponding x value in the exam!

(x) = sinxcos x + (cosx) ^ 2 = sin (2x) / 2 + [1 + cos (2x)] / 2 = (1 / 2) [sin (2x) + cos (2x)] + 1 / 2 = (1 / 2) [sin (2x) + cos (2x)] + 1 / 2 = (√ 2 / 2) sin (2x + π / 4) + 1 / 2 minimum positive period = 2 π / 2 = π sin (2x + π / 4) = 1, the function has the maximum value f (x) max = (√ 2 + 1) / 2) at this time, 2x + π / 4 = 2K π + π + π / 2 / 2 at this time, 2x + π / 4 = 2K π + π + π + π / 2 / 2, at this time, x = k π + π / 8, K

If 0 < x < π / 4, what is the minimum value of the function f (x) = (COS ^ 2x) / (sinxcosx sin ^ 2x)? The process should be detailed. If it is good, it will be added

Analysis:
Given that 0 < x < π / 4, then: 0

The known function f (x) = cos ^ 2 (x + 12 times π) - 1, G (x) = sinxcosx 1. Find the symmetry axis equation of the image with function y = f (x) 2. Find the value range of sum (x) = f (x) + G (x)

In this paper, K is an arbitrary integer. H (x) = f (x) = f (x) = [1 + cos (2x + π / 6)] / 2 symmetric axis is 2x + π / 6 = k π, that is, x = (k-1 / 6) π / 2, where k is any integer. H (x) = f (x) + G (x) = [1 + cos (2x + π / 6)] / 2 + 1 / 2 * sin2x = [1 + sin2x + cos2x] / 2 = [1 + √ 2Sin (2x + π / 4)] / 2 / 2, the maximum value is (1 + √ 2) / 2, the minimum value is (1 - - (1 - - (1 - - (2) / 2), the minimum value is (1 - - (1 - (1 - (1 - (√ 2) / 2 range

Let f (x) = cos ^ 2 (x + π / 12) + sinxcosx

f(x)=[1+cos(2x+π/6)]/2+(sin2x)/2=1/2+1/2[sin(π/3-2x)+sin2x]=1/2+sin(π/6)cos(π/6-2x)=1/2+1/2* cos(2x-π/6)cos(2x-π/6)=1,fmax=1/2+1/2=1cos(2x-π/6)=-1,fmin=1/2-1/2=0

Symmetric axis equation of F (x) = 2sinxcosx + 2cos ^ x-radical 3 + 2 Good guys, do it

f(x))=2sinxcosx+2√3cos^2x-3√3+2
=sin2x+√3cos2x+2
= 2sin(2x+π/3)+2
Thus, the symmetric axis equation x = k π / 2 + π / 12 (k is an integer)

It is known that the distance between the two adjacent symmetry axes of the image f (x) = cos ^ 2 ω x + 2 Radix 3cos ω x + sin ω x-sin ^ 2 ω x is g / 2 1: Find the value of ω 2: In the triangle ABC, a, B, C are the opposite sides of angles a, B and C. If a = √ 3, f (a) = 1, find the maximum value of B + C

Given the function f (x) = cos ^ 2 ω x + 2 √ 3cos ω x + sin ω x-sin ^ 2 ω x, the distance between the two adjacent symmetry axes of the image is π / 21: find the value of ω 2: in the triangle ABC, a, B, C are the opposite sides of angles a, B, C respectively, if a = √ 3, f (a) = 1, find the maximum value of B + C (1) analytic: ∵ function f (x) = cos ^ 2 ω x + 2 √ 3cos ω XS

The (π + s) root set of the equation

Sin (π + x) = root 3cos (π - x) gives - SiNx = - root 3cosx, that is, TaNx = radical 3, so x = k π + π / 3

Given the function f (x) = x half of COS (x of sin + 3 times of root of COS) - root of 2 / 3, find the equation of symmetric axis of function y = f (x)

(x) = cosx / 2 (SiNx / 2 + 3 + 3 cosx / 2) - √ 3 / 2 = SiNx / 2cosx / 2 + (3) 3 / 2 = 3 / 2 (1 + cosx) 3 / 2 = 1 / 2 SiNx + 3 / 2 (1 + cosx) 3 / 2 (1 + cosx) 3 / 2 (1 + cosx) 3 / 2 = SiNx cos π / 3 + cosxsin π / 3 = sin (x + π / 3) from x + π / 3 = k π + π / 2, K ∈ Z, x = k π + π / 6, K ∈ Z, that is, K ∈ Z, that is, K ∈ Z, that is, K ∈ Z, namely, K ∈ Z, namely, K ∈ z the symmetry axis of y = f (x)

The symmetry axis of the image of the function y = sin2x + cos2x is

y=sin(2x)+cos(2x)=√2*sin(2x+π/4) ,
From 2x + π / 4 = k π + π / 2, x = k π / 2 + π / 8,
Therefore, the symmetry axis of the function image is x = k π / 2 + π / 8, K ∈ Z