Find the period and monotone interval of the function f (x) = cos ^ 2x + 2 √ 3sinxcosx sin ^ 2x

Find the period and monotone interval of the function f (x) = cos ^ 2x + 2 √ 3sinxcosx sin ^ 2x

f(x)=cos^2x+2√3sinxcosx-sin^2x
=cos^2x-sin^2x+2√3sinxcosx
=cos2x+2√3sinxcosx
=cos2x+√3sin2x
=2[(1/2)*cos2x+(√3/2)*sin2x]
=2*sin（2x+π/6）
=2*sin[2(x+π/12)]
The period is 2 π / 2 = π
The monotonic interval is:
Monotonic increase on [K π - π / 3, K π + π / 6);
[K π + π / 6, K π + 2 π / 3)

Minimum positive period and maximum value of function f (x) = sin (2x + π / 6) + cos (2x + π / 3)

Sin (2x + π / 6) + cos (2x + π / 3) = √ 2Sin (2x + π / 6 + π / 4) = √ 2Sin (2x + 5 π / 12) (formula asinx + bcosx = √ (a ^ 2 + B ^ 2) sin (x + argtan (B / a)? T = 2 π / 2 = π, 2x + 5 π / 12 = π / 2 + 2K π, x = 1 / 24 π + K π, f (x) = √ 2

The minimum positive period and maximum of the function y = sin (2x + π / 6) - cos (2x + π / 3) are

y=sin(2x+π/6)-cos(2x+π/3)
=sin2x*(√3/2)+cos2x*(1/2)-cos2x*(1/2)+sin2x*(√3/2)
=√3sin2x
Then t = 2 π / 2 = π
The maximum value is √ 3
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Find the minimum positive period, maximum value and minimum value of the function y = cos ^ 2x sin ^ 2x

Y = cos ^ 2x sin ^ 2x = cos2x (according to cos2x = COS) ² x-sin ² x）
So the minimum positive period is t = 2 π / 2 = π
Maximum value: Ymin = - 1, minimum value: ymax = 1

Find the minimum period and maximum value of the function y = sin (2x + π / 6) - cos (2x + π / 3). Explain in detail,

y=sin(2x+π/6)-cos(2x+π/3)
=sin2xcosπ/6+cos2xsinπ/6-(cos2xcosπ/3-sin2xsinπ/3)
=√3sin2x
Minimum positive period π and maximum value √ 3

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