Trigonometric function FX = 2 √ 3cos ^ 2 + 2sinxcosx-m, in the interval [0] Trigonometric function function FX = 2 √ 3cos ^ 2 + 2sinxcosx-m, on the interval [0, π / 2], the maximum value of the function is 2, and find the value of M. in △ ABC, the opposite side of angle ABC is ABC. If a is an acute angle and satisfies FA = 0, SINB = 3sinc, △ ABC area is 3 √ 3 / 4, find the side length a

f(x)=√3cos2x+sin2x+√3-m=2sin(2x+pi/3)+√3-m max=2=2+√3-m m=√3
f(A)=0=2sin(2A+pi/3) A=pi/3 tanB=√3/5 ctanB=5/√3 sin^2B=1/(1+ctg^2B)=3/28
S=1/2bcsinA=1/2*(a/sinA)^2*sinB*sinC*sinA=a^2/√3*sinB^2/3=3√3/4 a=3√7

Given that f (x) = sin (2x + π / 3) + sin (2x - π / 3) + 2cosx square, X ∈ R (1) find the minimum positive period of F (x) (2) find the monotone decreasing interval of F (x) (3) if the function g (x) = f (x) - M has no zero point on the interval (- π / 4, π / 4), find the value range of M

sin(2x+π/3)=sin2xcos60+sin60cos2xsin(2x-π/3)=sin2xcos60-sin60cos2x2(cosx)^2=2×[(1+cos2x)/2]=cos2x+1f(x)=2sin2xcos60+cos2x+1f(x)=sin2x+cos2x+1sin2x＋cos2x=√2(√2/2sin2x+√2/2cos2x)=√2(sin2xcos45+c...

F (x) = sin (2x + π / 3) + √ 3cos (2x + π / 3), the period of F (x), monotone decreasing interval Why?

f(x)=2sin(2x+π/3+π/3)
T=2π/2=π
Simple subtraction (K π - 1 / 12 π, K π + 5 / 12 π)

Finding the monotone decreasing interval of function y = cos (PAI / 3-2x)

COS is even function
So y = cos (2x - π / 3)
The cosx minus interval is (2k π, 2K π + π)
So 2K π < 2x - π / 3 < 2K π + π
2kπ+π/3<2x<2kπ+4π/3
K π + π / 6 is (K π + π / 6, K π + 2 π / 3)

The function y = sin (π 4-2x)______ .

The function y = sin (π 4-2x) = - sin (2x - π 4) because π 2 + 2K π ≤ 2x − π 4 ≤ 3 π 2 + 2K π K ∈ Z, the solution is: 3 π 8 + K π ≤ x ≤ 7 π 8 + K π K ∈ Z, so the increasing interval of function y = sin (π 4-2x) is 3 π 8 + K π ≤ x ≤ 7 π 8 + K π (K ∈ z)

Let y = sin (PAI / 3-2x) be known to find the period, monotone decreasing interval, symmetry axis, center of symmetry, maximum value and the set of X with maximum value

The minimum positive period is π,
The monotone decreasing interval is [- π / 12 + K π, 5 π / 12 + K π], and K is an integer
The axis of symmetry is x = - π / 12 + K π, and K is an integer
The center of symmetry is (π / 6 + K π, 0) k is an integer
The maximum value is 1, where x = - π / 12 + K π
The minimum value is - 1, where x = 5 π / 12 + K π

Monotone decreasing interval and maximum value of y = - sin (2x Pai / 6) + 3 / 2

When y = - sin (2x Pai / 6) + 3 / 2 decreases, 2x Pai / 6 belongs to 2K "Pai" - 0.5 "Pai" to 2K "Pai" + 0.5 "Pai"
When k "Pai" + 0.5 "Pai", the maximum value is 2x Pai / 6

F (x) = cos (2x + Pai / 4) + sin (2x + Pai / 4) In my preview, I didn't see this kind of analysis about the combination of two trigonometric functions to find the monotone interval,

cos(2x+pai/4)+sin(2x+pai/4)=√2*[√2/2cos(2x+pai/4)+√2/2sin(2x+pai/4)]
=√2*[sin45*cos(2x+pai/4)+cos45sin(2x+pai/4)]=√2sin(45+2x+45)=√2sin(x+90)=-√2cos2x
[k*pai,(k+1/2)*pai]

The monotone decreasing interval of the function f (x) = sin (2 x divided by 6) is?

f(x)=-sin(2x-π/6)
The decreasing interval of F (x) is the increasing range of sin (2x - π / 6)
That is 2x - π / 6 ∈ [- π / 2 + 2K π, π / 2 + 2K π]
x∈[-π/6+kπ,π/3+kπ],k∈Z

Find the function y = sin (- 2x + π) 6) The monotone decreasing interval, the maximum value and the value set of X when taking the maximum value

The function y = sin (- 2x + π)
6）=-sin（2x-π
6) The monotone decreasing interval of,
That is, the function T = sin (2x - π)
6) Monotonically increasing interval of
Let 2K π - π
2≤2x-π
6≤2kπ+π
2, K ∈ Z, find K π - π
6≤x≤kπ+π
3，
Therefore, the decreasing interval of the function is [− π
6+kπ，π
3+kπ](k∈Z)．
When 2x - π
6=2kπ-π
2, K ∈ Z, that is, x = k π − π
At 6, the maximum value of the function is 1;
When 2x - π
6=2kπ+π
2, K ∈ Z, that is, x = k π + π
When 3, the minimum value of the function is - 1