Given the function f (x) = radical 3sin2x-2cos ^ 2x-1, X ∈ R, in △ ABC, the opposite sides of a, B and C are a, B, C, C = radical 3, f (c) = 0 SINB = 2sina, find the value of a, B

Given the function f (x) = radical 3sin2x-2cos ^ 2x-1, X ∈ R, in △ ABC, the opposite sides of a, B and C are a, B, C, C = radical 3, f (c) = 0 SINB = 2sina, find the value of a, B

The number of root is 3 / 2sin2x-1 / 2cosx ^ 2-1-1 = 3-sin2x-cos2x-2 = 2 (root 3 / 2sin2x-1 / 2cos2x) - 2 = 2Sin (2x - π / 6) - 2 = 2Sin (2x - π / 6) - 2 because f (c) = 0, so f (c) = 2Sin (2C - π / 6) - 2 = 02sin (2C - π / 6) - 2 = 02sin (2C - π / 6) = 2, namely sin (2C - π / 6 / 6) = 1, so 2C - π / 6 = π / 2 = π / 2, C = π / 3, because because f (c) = 0, because f (C = C = C = π / 3, because f (c) = 0, so 2C - π SINB = 2Si

Find a symmetric center point of the function f (x) = [radical 2Sin (x + π / 4) + x] / cosx

Find a symmetric center point of the function f (x) = [radical 2Sin (x + π / 4) + x] / cosx
Analysis: ∵ function f (x) = [√ 2Sin (x + π / 4) + x] / cosx
Let f '(x) = √ 2cos (x + π / 4) cosx + cosx + √ 2Sin (x + π / 4) SiNx + xsinx / (cosx) ^ 2
=[√2cos(π/4)+cosx+xsinx]/(cosx)^2=(1+cosx+xsinx)/(cosx)^2=0
It is found that X1 = - π, X2 = π
F(X1)=π+1,F(X2)=1-π,F(0)=1
It can be conjectured that a symmetric central point of the function f (x) = [√ 2Sin (x + π / 4) + x] / cosx is a point (0,1)

Find the definition domain of radical 2 × cosx-1 under the function f (x) = lgsinx +

2×cosx-1≥0 ==> cosx≥1/2 ===> -π/3+2kπ≤x≤π/3+2kπ,k∈Z (1)
sinx>0 ==> 2kπ

Let f (x) = 1-radical 2Sin (2X-4) divide cosx to find the domain of F (x)?

Since your description is rather complicated, I tried to analyze your topic. Do you think it is right? Find the definition domain of F (x) = [1 - √ 2Sin (2x - π / 4)] / cosx. If so, the solution is as follows: in order to make the function meaningful, sin (2x - π / 4) ≥ 0 and cosx ≠ 0, so 2K π ≤ 2x - π / 4 ≤ 2K π + π

Given the function f (x) = (1-radical 2Sin (2x-45 degrees)) divided by cosx, find the domain of F (x)

K π + π / 8 ≤ x < π / 2 or π / 2 < x ≤ 5 π / 8 + K π
2Sin (2x-45 degree) ≥ 0 K π + π / 8 ≤ x ≤ 5 π / 8 + K π
cosx≠0 x≠π/2
Take the intersection to get k π + π / 8 ≤ x < π / 2 or π / 2 < x ≤ 5 π / 8 + K π

（2cos^2-1）/2tan(π/4-α)sin^2(π/4+α)

（2cos^2α-1）/2tan(π/4-α)sin^2(π/4+α)
=（1+cos2α-1）/[2tan(π/4-α)sin^2(π/4+α)]
=cos2x/[2tan(π/4-α)sin^2(π/4+α)]
2tan(π/4-α)sin^2(π/4+α)=[2(sin(π/2-2α))/(cos(π/2-2α)+1)]x[(1-cos(2α+π/2))/2]
=[2(cos2α/(sin2α+1)]*[(1+sin2α)/2]
=cos2x

Calculation: 2cos2 α - 1 2tan (π 4−α)•sin2(π 4+α)．

Original formula = Cos2 α
2sin(π
4−α)
cos(π
4−α)•cos2(π
4−α)=cos2α
2sin(π
4−α)cos(π
4−α)＝cos2α
sin(π
2−2α)=cos2α
cos2α=1

(2cos^2a-1)/(2tan(pai/4+a)sin^2(pai/4-a))

(2cos^2a-1)/(2tan(pai/4+a)sin^2(pai/4-a))
= (cos2a)/{[2sin(π/4+a)/cos(π/4+a)]*sin²(π/4-a)}
∵ sin(π/4+a)=sin[(π/2)-(π/4-a)]=sin(π/4-a)
cos(π/4+a)=cos[(π/2)-(π/4-a)]=sin(π/4-a)
= (cos2a)/{[2cos(π/4-a)/sin(π/4-a)]*sin²(π/4-a)}
=(cos2a)/[2cos(π/4-a)*sin(π/4-a)]
=cos2a/sin(π/2-2a)
=cos2a/(cos2a)
=1

Given the function f (x) = 2acos ^ 2x + bsinxcosx-1, if f (0) = 1, f (π / 3) = - 1 / 2 + radical 3 / 2, if f (x) > 1, find the value range of X

f(x)=2acos²x+bsinxcosx-1
f(0)=2acos²(0)+bsin0cos0-1=1
2a-1=1 a=1
f(π/3)=2cos²(π/3)+bsin(π/3)cos(π/3)-1=-1/2+√3/2
1/2+b(√3/2)(1/2)=1/2+√3/2
B=2
f(x)=2cos²(x)+2sinxcosx-1
=cos(2x)+sin(2x)
=sin(π/2-2x)+sin(2x)
=2sin(π/4)cos(π/4-2x)
=√2cos(π/4-2x)>1
cos(π/4-2x)>√2/2
-π/4+2kπ

Given the function f (x) = x + 1 under the radical sign, (1) find the value of F (3) and (2) find the definition domain of F (x)

One
f(3)=√(3+1)=√4=2
Two
(x+1)>=0
x>=-1