Let the function f (x) = cos (2x + π) 3)+sin2x， (1) Find the maximum value and minimum positive period of function f (x); (2) Solve the trigonometric equation: F (x) = 0

Let the function f (x) = cos (2x + π) 3)+sin2x， (1) Find the maximum value and minimum positive period of function f (x); (2) Solve the trigonometric equation: F (x) = 0

（1）f（x）=cos（2x+π
3）+sin2x=cos2xcosπ
3−sin2xsinπ
3+1−cos2x
2＝1
2−
three
2sin2x
So the maximum value of function f (x) is 1+
three
2. Minimum positive period π
(2) From F (x) = 0, we get    one
2−
three
2sin2x=0    That is, sin2x =
three
3. Get   x＝1
2[kπ+(−1)karcsin
three
3]，x∈N

Function y = sin (2x + π) 6）+cos（2x+π 3) The minimum positive period and maximum value of are () A. π， two B. π，1 C. 2π， two D. 2π，1

∵y=sin（2x+π
6）+cos（2x+π
3）=
three
2sin2x+1
2cos2x+（1
2cos2x-
three
2sin2x）=cos2x，
The minimum positive period T = 2 π
2=π，ymax=1．
Therefore: B

Let the function f (x) = cos (2x + π 3) + sin square x, ask the maximum and minimum positive period of the function f (x)

F (x) = cos (2x + π / 3) + sin square x
=1/2cos2x-√3/2sin2x+(1-cos2x)/2
=-√3/2sin2x+1/2
Maximum value of F (x) = √ 3 / 2 + 1 / 2
Minimum positive period T = π

Let f (x) = cos (2x + π / 3) + sin ² x. Finding the minimum positive period of a function Want process

f(x)=cos(2x+π/3)+(sinx)^2
=(cos2x)/2+√3(sin2x)/2+(1-cos2x)/2 (∵cos2x=1-2(sinx)^2,∴(sinx)^2=(1-cos2x)/2)
=1/2+√3(sin2x)/2
The minimum positive period of the function is π

Y = derivative of COS [㏑ (1 + 2x)],

y '=-sin[ln(1+2x)] × [ln(1+2x)] '
=-sin[ln(1+2x)] × 1/(1+2x) × (1+2x) '
=-sin[ln(1+2x)] × 1/(1+2x) × two
=-2sin[ln(1+2x)]/(1+2x)

Y = derivative of COS (3x-1) - ln (- 2x-1)

-3SIN(3X-1)+2/(-2X-1)