Calculus

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

Well,

you could just substitute each of those values into the left hand side and see which one gives and answer closest to 6.

Very nicely done Omar :)