The function f (x) = 2sinxcosx + cos2x (x) is known

The function f (x) = 2sinxcosx + cos2x (x) is known

f(x)=sin2x+cos2x
=√2sin(2x+π/4)
So t = 2 π / 2 = π
Maximum = √ 2
f(θ+π/8)=√2sin(2θ+π/4+π/4)
=√2cos2θ
=√2/3
cos2θ=1/3
When the angle of θ is acute, sin2 θ > 0
sin²2θ+cos²2θ=1
sin2θ=2√2/3
tan2θ=2√2

(2005 · Anhui) when 0 < x < π When 2, the function f (x) = 1 + cos2x + 8sin2x The minimum value of sin2x is () A. 2 B. 2 Three C. 4 D. 4 Three

f(x)＝2cos2x+8sin2x
2sinxcosx＝4sin2x+cos2x
sinxcosx=4tan2x+1
tanx＝4tanx+1
tanx．
∵0＜x＜π
2，
∴tanx＞0．
∴4tanx+1
tanx≥2
4tanx•1
tanx＝4．
When TaNx = 1
2, f (x) min = 4
Therefore, C

Let f (x) = a * B, where the vector a = (2cosx, 1), B = (cosx, root 3 * sin2x), X belongs to R. (1) find the minimum positive period of F (x); (2) in △

2X + s 2x (2x) + 2x = 2x (2x) + 2x = 2x (2x) + 2x (2x) + 2x
The minimum positive period of F (x) t = 2 π / 2 = π

Find the maximum and minimum values of the function y = (√ 3 / 2) sin2x + (cosx) ^ 2 and the set of X whose maximum and minimum y are obtained

y=(√3/2)sin2x+1/2(cos2x+1)=(√3/2)sin2x+(1/2)cos2x+1/2=sin(2x+π/6)+1/2
sin(2x+π/6)∈【-1,1】
So y Max is 3 / 2 and min is - 1 / 2
The period of sin (2x + π / 6) is π
When the minimum value is x = - π / 3 + K π = - π / 3 + 2K (π / 2) (K ∈ z)
When the maximum value is x = π / 6 + K π = - π / 3 + (2k + 1) (π / 2) (K ∈ z)
So {x | x = - π / 3 + n · π / 2, n ∈ Z}

Given f (x) = sin2x-2cos ^ 2x, find the minimum positive period of function f (x) Big brother and elder sister help. I'm stupid and can't do it

f(x)=sin2x-2cos^2x
=sin2x-cos2x-1
=√2(sin2x-π/4)-1
∴T=2π/2=π
The minimum positive period of function f (x) is π

Find the minimum positive period of the function f (x) = sin2x-2cos ^ 2x

f(x)=sin2x-(cos2x+1)
=sin2x-cos2x-1
=√2sin(2x-П/4)-1
Therefore, the minimum positive period T = 2 П / 2 = П

The known function f (x) = 2cos square x-1) sin 2x + 1 / 2 cos4x Find the minimum positive period of FX) and its maximum value If a belongs to (2 parts of π, π) and f (a) = the root of 2, find the value of A

The solution of the solution f (x) = (2cos? - 1) sin2x + 1 / 2cos4x = cos2xsin2x + 1 / 2cos4x = 1 / 2sin4x + 1 / 2cos4x = 1 / 2sin4x + 1 / 2cos4x = 1 / 2sin4x + 1 / 2cos4x = √ 2 / 2Sin (4x + π / 4) / / T = 2 π / 4 = π / 2 is its minimum cycle, the maximum value of its minimum cycle is: √ 2 / 2F (a) = √ 2 / 2F (a) = √ 2 / 2 ͪ 4 ͪ 4 ͪ 4 ͪ 4 = π / 2 + 2 + 2K π π π π (π / 16 / 16 / 16 / + K π / 2 ∵ a ∈ (...)

Given the function f (x) = (sin2x / SiNx) + 2Sin, find the minimum positive period of function f (x)?

f(x)=(2sinxcosx/sinx)+2sinx=2(cosx+sinx)=2√2[sin(π/4)cosx+cos(π/4)sinx]=2(√2)sin(x+π/4)
Therefore, the minimum positive period T = 2 π

The minimum positive period of the function f (x) = sin2x + radical 3cos2x is Asinx + bcosx general extraction root (a square + b square) So this problem is extracted from 2 2 (1 / 2sin2x + Radix 3 / 2cos2x) = 2Sin (2x + 60 °) So the period is 2 π - 2 = π How did 2Sin (2x + 60 °) come from?

(1 / 2sin2x + Radix 3 / 2cos2x) = sin2xcos60 + cos2xsin60 = sin (2x + 60 °)
formula

The function f (x) = sin2x+ The minimum positive period of 3cos2x is______ .

It can be concluded from the meaning of the title
y=sin2x+
3cos2x
=2（ 1
2sin2x+
Three
2cos2x）
=2sin（2x+π
3）
∴T=2π
2=π
So the answer is: π