Let the function fx root 3 coswx^2 sinwx coswx a where w 0

Suppose the function fx has the root number 3 coswx^2 sinwx coswx a where w 0 a belongs to R

f(x)=√3(coswx)^2 sinwxcoswx a

=root 3 (cos2wx 1)/2 sin2wx/2 a

= sin(2wx π/3) √3 / 2 a,

The abscissa of the first lowest point on the right side of the y-axis of the image of f(x) is 7π/6.

So when x=7π/6, 2w*7π/6 π/3=3π/2,

W=1/2.

So f(x)= sin(x π/3) √3 / 2 a,

X∈[-π/3,5π/6], then x π/3∈[0,7π/6],

The minimum value of sin(x π/3) is sin7π/6=-1/2,

The minimum value of sin(x π/3) √3 / 2 a is -1/2 √3 / 2 a,

So -1/2 √3 / 2 a=√3,

a=（√3 1）/2.

Suppose the root of the function is 3sinxcosx cos^2 x R

y=root number 3sinxcox cos^2x

=root number 3sinxcox (1/2-1/2cos2x)

=(root 3/2) sin2x 1/2-1/2co2x

=sin2xcospie/6-cos2xsinpie/6 1/2

=sin(2x-pie/6) 1/2

-pie/3-2pie/3-pie/2 and because sin(2x-pie/6) 1/2=-root number 3/2 1/2

sin(2x-pie/6)=-root number 3/2

Because sin (pie/2-pie/6)=cos pie/6=-cos root number 3/2

2x Pie/6=2/ Pie Pie/2 Pie/6

x=pie/4

y=sin(2x-pie/6) 1/2 abscissa is compressed to the original 1/2

g(x)=(4x-PI/6) 1/2 translation ∏/6 units, and finally translation upwards 1 unit

y=sin(4x-5/6 faction) 3/2 interval, write it yourself

Suppose the function fx root 3cos squared ωx sinωx a

Explanation "√3" means root number 3, pai is π

Solution: (1) Let t=sinwx, then y=-√3t^2 t (√3-a) According to the meaning of the question: t=sinwx image passes the origin and pai/6 is the first maximum value point

So, the period of t=sinwx is T=(pai/6)*4=(2pai)/3

According to the period formula: T=(2pai)/3=(2pai)/w, so w=3

(2)For t=sin3x,-paiy=-√3t^2 t (√3-a) axis of symmetry t=1/(2√3)

y=-√3t^2 t (√3-a) first increases and then decreases on [-1,1], so the minimum value of y is y(-1) or y(1)

And y(-1)=-√3 (-1) √3-a=-1-a ; y(1)=-√3 1 √3-a=1-a then y(min)= -1-a=√3

a=-1-√3

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