Let f (x) = 2cos square x + sin2x + a a a belong to R. when x belongs to [0, π / 6], the maximum value of F (x) is 2, and the value of a is obtained, The symmetric axis equation of y = f (x) x belonging to R is obtained Don't give the teacher an answer

Let f (x) = 2cos square x + sin2x + a a a belong to R. when x belongs to [0, π / 6], the maximum value of F (x) is 2, and the value of a is obtained, The symmetric axis equation of y = f (x) x belonging to R is obtained Don't give the teacher an answer

It is agreed that% is the sign of square root, such as 2% = root 2
F (x) = 2cos square x + sin2x + A
f(x)=cos2x+sin2x+a+1
f(x)=2%*（2%/2cos2x+2%/2sin2x）+a+1
f(x)=2%*sin(2x+π/4)+a+1
When 2x = π / 4, f (x) is the maximum at [0, π / 6]
That is, f (π / 4) = 2
A = 1-2%
Find the axis of symmetry:
Let 2x + π / 4 = π / 2 or 2x + π / 4 = 3 π / 2, and f (x) be a periodic function with period π
So the equation of symmetry axis of F (x) is
X = π / 8 + K π or x = 3 π / 8 + K π
K is an integer
Remember to add points to say

Given the function f (x) = sin2x + 2cos ^ 2x-1 (1), find the minimum positive period and maximum value of function f (x) (2) find the function in the interval (2) Find the maximum and minimum value of function in the interval [π / 4,3 π / 4]

f(x)=sin2x+cos2x=√2sin(2x+π/4)
（1）T=2π/2=π,f(x)max=√2；
（2）x∈【π/4,3π/4】
Then: 2x + π / 4 ∈ [3 π / 4,7 π / 4]
Then: sin (2x + π / 4) ∈ [- 1, √ 2 / 2]
Therefore, f (x) = √ 2Sin (2x + π / 4) ∈ [- √ 2,1]
In other words, the maximum value of the function in the interval [π / 4,3 π / 4] is 1 and the minimum value is - √ 2

Find the maximum value of the function y = 2cos ^ 2x + sin2x-2

Solution y = 2cos ^ 2x + sin2x-2
=2cos^2x-1+sin2x-1
=cos2x+sin2x-1
=√2（√2/2sin2x+√2/2cos2x）-1
=√2sin（2x+π/4)-1
≤√2-1
Therefore, the maximum value of y = 2cos ^ 2x + sin2x-2 is √ 2-1

Function y = sin2x-2cos ^ 2x max? How much is it?

The maximum value is (root 2) - 1

Given the function f (x) = sin2x + 2cos (pai-x) cosx + 1 (1), find the minimum positive period and maximum value of the function (2) if the triangle ABC three sides ABC corresponds to the angle ABC and satisfies the square of B + the square of C = the square of a + √ 2 BC, find the value of F (a)

f(x)=sin2x+2cos(π-x)cosx+1
=sin2x-2cos^2x+1
=sin2x-cos2x
=√2sin(2x-π/4)
The minimum positive period T = π, the maximum y = √ 2
b^2+c^2=a^2+√2bc,a^2=b^2+c^2-2bccosA,a^2+√2bc=a^2-2bccosA,cosA=-√2/2,A=3π/4
f(3π/4)=√2sin(3π/2-π/4)=√2sin5π/4=-1

Y = 2sinxcosx = sin2x, so the minimum period is π! What is the function and why?

This is a trigonometric function
T (period) = 2 π △ ω
Here Omega is two,
So t = π
In addition, sin2x = 2sinxcosx is one of the trigonometric double angle formulas, which can be used directly

What is the minimum positive period of the function y = cosx ^ 2-2sinxcosx-sin2x What is the domain of y = LG (x ^ 2 + X-2)?

y=(cos2x +1)/2-sin2x-sin2x=cos2x/2 -2sin2x+1/2
T = pie
X ^ 2 + X-2 > 0 x < - 2 or x > 1

The minimum positive period of the function y = 2 * sin ^ 2 x - sin2x - 1 is

y=2*sin^2 x-sin2x-1
=-cos2x-sin2x
=-√2sin(2x+π/4)
T=2π/2=π

1. The minimum positive period of function y equal to sin2x is 2. The maximum value of function y equal to cos3x minus root sign 3sin3x is

y=sin2x
T=2π/2=π
y=-2(sin3x*√3/2-cos3x*1/2)
=-2(sin3xcosπ/6-cos3xsinπ/6)
=-2sin(3x-π/6)
So the maximum value is 2

The minimum positive period of the function y = 1 / 2 | sin2x | is

The period of sin 2x is 2 π / 2 = π
With the absolute value added, the period is halved, which is π / 2