It is known that f (x) = 2Sin (x + θ / 2) cos (x + θ / 2) + 2 √ 3cos ^ 2 (x + θ / 2) -√ 3 Question: if 0 ≤ θ ≤ π, and the function f (x) is even, find the set of X satisfying the equation f (x) = 1 and X ∈ [0, π] The answer is x = 5 π / 6 or x = π / 6 I just can't figure out pi / 6

f（x）=2sin（x+θ/2）cos（x+θ/2）+2√3cos^2（x+θ/2）-√3
=sin(2x+θ)+√3(1+cos(2x+θ))-√3
=sin(2x+θ)+√3cos(2x+θ)
=2sin(2x+θ+π/3)
F (x) is even function
Then (π + π / k)
And 0 ≤ θ ≤ π
Then θ = π / 6
So f (x) = 2Sin (2x + π / 6 + π / 3) = 2cos2x
Let f (x) = 1 to get 2cos2x = 1
So cos2x = 1 / 2
So 2x = 2K π ± π / 3 (K ∈ z)
That is, x = k π ± π / 6 (K ∈ z)
And X ∈ [0, π]
Then x = π / 6 or 5 π / 6
If you do not understand, please hi me, I wish you a happy study!

Given Tana = - 1 / 3, cos β = √ 5 / 5, a, β∈ (0, π), find the maximum value of the function f (x) = √ 2Sin (x-a) + cos (x + β)

∵tana=-1/3 a∈(0,π),
∴sina/cosa=-1/3,cosa=-3sina
Substituting sin? A + cos? A = 1 = = > sin? A = 1 / 10
∴,sina=√10/10,cosa=-3√10/10
∵cosβ=√5/5,a,β∈(0,π),
∴sinβ=2√5/5
∴f(x)=√2sin(x-a)+cos(x+β)
=√2(sinxcosa-cosxsina)+cosxcosβ-sinxsinβ
= -3√5/5sinx-√5/5cosx+√5/5cosx-2√5/5sinx
=-√5sinx
The maximum value of F (x) is √ 5

F (x) = 2cos ^ 2 x + 2Sin x cos x

Using the formula of double angle
2sinxcosx=sin2x
cos2x=2(cosx)^2-1
cos2x+1=2(cosx)^2
f(x)=cos2x+1+sin2x
=√2sin(2x+π/4)+1
Maximum √ 2 + 1
Minimum - √ 2 + 1

How to find the period of function 2Sin (x / 3) + cos (x / 2)?

The period of sin (x / 3) is 2 π / (1 / 3) = 6 π
The period of COS (x / 2) is 2 π / (1 / 2) = 4 π
Because the least common multiple of 4 and 6 is 12, the period of 2Sin (x / 3) + cos (x / 2) is 12 π

1、 21. The period of the function cos (x / 4) - 2Sin (x / 3) is 11000 your answer is too simple, and the answer is not right, I want a detailed calculation process

The period of COS (x / 4) is 2 π / (1 / 4) = 8 π, and the period of sin (x / 3) is 2 π / (1 / 3) = 6 π, so the common period is 24 π, that is, the period of COS (x / 4) - 2Sin (x / 3) is 24 π (this kind of problem can only be dealt with in this way, if necessary, it can be proved according to the definition of period)

Find the period of COS (x / 2) + 2Sin (x / 3) Please note the process, thank you

The period of COS (x / 2) is 2 π / (1 / 2) = 4 π
The period of sin (x / 3) is 2 π / (1 / 3) = 6 π
So the period of the function is the least common multiple * π of 4 and 6
So period T = 12 π

It is known that f (x) = sin2x − 2sin2x 1−tanx． (I) find the definition domain and the minimum positive period of the function f (x); (II) when cos (π) 4+x)＝3 5, find the value of F (x)

X ≠ K π + π 4 (K ∈ z) and X ≠ K π + π 4 (K ∈ z) from 1-tanx ≠ 0. And X ≠ K π + π 2 (K ∈ z) 2 (K ∈ z) function, X ≠ K π + π 4, X ≠ K π + π 4, X ≠ K π + π + π 2 (K ∈ z)} ∵ f (x) = sin2x − 2sin2x1 − TaNx = cosx · 2sinx (cosx − SiNx) cosx − cosx − cosx − SiNx (cosx − SiNx) cosx − cosx − cosx SiNx = sin2x, ν the minimum of F (x)

1, f (x) = cos (2x - π / 3) + 2Sin ^ 2 x (1) to find the equation of the least positive period and the axis of symmetry,

F (x) = cos (2x - π / 3) + 2Sin ^ 2 x = cos2xcos π / 3 + sin2xsin π / 3 + 1-cos2x = 1 / 2cos2x + sin2xsin π / 3 + 1-cos2x = - 1 / 2cos2x + sin2xsin (π - 2 π / 3) + 1 = Cos2 π / 3cos2x + sin2xsin 2 π / 3 + 1 = cos (2x-2 π / 3) + 1, so the minimum positive period is π, and the equation of symmetry axis is x =

Verification: （1）2sin(π+θ)•cosθ−1 1−2sin2θ＝tan(9 π+θ)+1 tan(π+θ)−1；（2）tanθ•sinθ tanθ−sinθ＝cosθ•(tanθ+sinθ) sin2θ．

It is proved that: (1) left = - 2Sin θ cos θ − 1
cos2θ−sin2θ=−(sinθ+cosθ)2
(sinθ+cosθ)cosθ−sinθ)＝(sinθ+cosθ)
(sinθ−cosθ)＝tanθ+1
tanθ−1=−sinθ−cosθ
cosθ−sinθ=−tanθ−1
1−tanθ=tanθ+1
tanθ−1；
Right = Tan (8 π + π + θ) + 1
tanθ−1=tanθ+1
tanθ−1，
 left = right, get the certificate;
(2) Left = sin θ
cosθ•sinθ
sinθ
cosθ−sinθ=sin2θ
sinθ(1−cosθ)=sinθ
1−cosθ，
Right = cos θ · (sin θ)
cosθ+sinθ)
sin2θ=sinθ(1+cosθ)
1−cos2θ=sinθ
1−cosθ，
﹤ left = right, get the certificate

It is proved that 2Sin (П + θ) cos θ - 1 / 1-2sin ^ 2 θ = Tan (9 П + θ) - 1 / Tan (П + θ) + 1

prove:
Left = 2Sin (П + θ) cos θ - 1 / 1-2sin ^ 2 θ
=(-2sinθcosθ-1)/cos2θ
=-(2sinθcosθ+sin^2 θ+cos^2 θ)/(cos^2 θ-sin^2 θ)
=-(sinθ+cosθ)^2/(cosθ-sinθ)(cosθ+sinθ)
=-(sinθ+cosθ)/(cosθ-sinθ)
=-[(sinθ/cosθ)+1]/[1-(sinθ/cosθ)]
=-(tanθ+1)/(1-tanθ)
=(tanθ+1)/(tanθ-1)
Right = Tan (9 П + θ) - 1 / Tan (П + θ) + 1
=(tanθ-1)/(tanθ+1)