pro35hp

Well , Melody's answer is correct if you don't need to proove it 

but if you need proof

\(\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides} \)

\(6\sin \left(2x\right)-8\cos \left(x\right)+9\sin \left(x\right)-6=0\)

\(Use \: \sin \left(2x\right)=2\cos \left(x\right)\sin \left(x\right)\)

\(-6-8\cos \left(x\right)+9\sin \left(x\right)+2\cdot \:6\cos \left(x\right)\sin \left(x\right)=0\)

factoring gives us \(\left(4\cos \left(x\right)+3\right)\left(3\sin \left(x\right)-2\right)=0\)

so either the first term or the second equivalent to 0 , we will check them both

\(4\cos \left(x\right)+3=0,\:\frac{-\pi }{2}\le \:x\le \frac{\pi }{2}\)

there are no solution in this range for x , you can check yourself

\(\sin \left(x\right)=\frac{2}{3}\)

in the range \(\frac{-\pi }{2}\le \:x\le \frac{\pi }{2}\)

there is only one solution for sinx= 2/3 in the range -pi/2 to pi/2

it is 

d. x = 0.730

\(\mathrm{Use\:the\:following\:identity}:\quad \sin \left(2x\right)=2\cos \left(x\right)\sin \left(x\right)\)

\(1.5\cos \left(x\right)+2\cos \left(x\right)\sin \left(x\right)=0\)

\(=\cos \left(x\right)\left(2\sin \left(x\right)+1.5\right)\)

\(\cos \left(x\right)=0\:\:\:\mathrm{or}\:\:\:\:2\sin \left(x\right)+1.5=0\)

\(\cos \left(x\right)=0\quad :\quad x=\frac{\pi }{2}+2\pi n,\:x=\frac{3\pi }{2}+2\pi n\)

\(2\sin \left(x\right)+1.5=0\quad :\quad x=2\pi n-0.84806,\:x=2\pi n+\pi +0.84806\)

\(x=2\pi n+\pi +0.84806,\:x=2\pi n-0.84806,\:x=\frac{3\pi }{2}+2\pi n,\:x=\frac{\pi }{2}+2\pi n\)

solve each x between 0 and 2pi

\(0<2\pi n-0.84806<2\pi\)

\(\frac{0.84806}{2pi}<n<1+ \frac{0.84806}{2pi }\)

hence , n = 1 , plug it in the previous equation \(2\pi -0.84806=5.435\)

So without calculating other equation

we have choices

and the true choice is 

e. x = 1.571, 3.990, 4.712, 5.435 

if you want to complete just solve each other x between 0 and 2 pi and don't forget that n is an integer number

sorry dear , but if it was that easy , it will be published as a formula or something like this

there are officials calculations

http://www.jstor.org/pss/2688509

http://www2.stetson.edu/~efriedma/cirincir/

http://en.wikipedia.org/wiki/Circle_packing_in_a_circle

but there is no such formula

as R2^2/r1^2 = number of circles of inside a R2 radius circle ...

cordially sir,

This is a hard problem , there are only some approximations

like \(\frac{π}{12}\times \left ( \frac{{R}_{2}^2}{{r}_{1}^2} \right ) - \frac{π}{12} \times \left ( \frac{{R}_{2}}{{r}_{1}} \right )\)

R2 is the large circle , r1 is the small circle

you can use an approximiate calculator 

http://www.engineeringtoolbox.com/smaller-circles-in-larger-circle-d_1849.html

You may find this interesting

http://www.jstor.org/pss/2688509

http://www2.stetson.edu/~efriedma/cirincir/

http://en.wikipedia.org/wiki/Circle_packing_in_a_circle

I think they're only using charts in order to define Cl

it called somewhat ( Velocity Relationship Curve Graph ) or something like this

not sure 

sorry

infinitely many equations

, remember any equations is defined by two points

here is only one point , you can give any other point as you like

and try to find the equation using \(m=\frac{y1-y2}{x1-x2}\)

and when you find m , try to find p , using either the first point or the point you guessed lol

\(\frac{d^3}{d^5} = d^3 * d^{-5} = d^{3-5} = d^{-2} \)

There are infinitely many solutions for this equation

you just need to give a value to x , in order to find y , or the opposite , giving a value to y in order to find x.