Oh my sun sin!

Equipped with the knowledge that this is an identity (since there is only one possible answer for $x$ ) we simply plug in a value for $A$ and solve for $x$ .

We plug in $x=30^{\circ}$ to get $\sin30\sin30\sin90=\dfrac{1}{x}\sin90$

Solving gives $\dfrac{1}{x}=\dfrac{1}{4}\implies x=4$

Thus the answer is $1.5\times 4=\boxed{6}$

$\begin{matrix} \sin\ { (60-A) } =\frac { \sqrt { 3 } }{ 2 } \cos\ { A } -\frac { 1 }{ 2 } \sin\ { A } \\ \sin\ { (60+A) } =\frac { \sqrt { 3 } }{ 2 } \cos\ { A }+\frac { 1 }{ 2 } \sin\ { A } \\ then, \\ \sin\ { A }.\sin\ { (60-A) } .\sin\ { (60+A) } = \\ \sin\ { A }.\left[ \frac { \sqrt { 3 } }{ 2 } \cos\ { A }-\frac { 1 }{ 2 } \sin\ { A } \right] .\left[ \frac { \sqrt { 3 } }{ 2 } \cos\ { A } +\frac { 1 }{ 2 } \sin\ { A } \right] \\ =\sin\ { A } .\left[ \frac { 3 }{ 4 } \cos\ ^{ 2 }{ A } -\frac { 1 }{ 4 } \sin\ ^{ 2 }{ A } \right] \\ but, \\ \cos\ ^{ 2 }{ A } =\frac { 1 }{ 2 } \quad \left[ 1\quad +\quad \cos\ { A } \right] \\ \sin\ ^{ 2 }{ A } =\frac { 1 }{ 2 } \quad \left[ 1\quad -\quad \cos\ { A } \right] \\ then, \\ =\sin\ { A }.\frac { 3 }{ 8 } (1+\cos\ { 2 }{ A })\quad -\quad \frac { 1 }{ 8 } (1-\cos\ { 2 }{ A }) \\ =\sin\ { A } .\left[ \frac { 3 }{ 8 } +\frac { 3 }{ 8 } \cos\ { 2 } { A } \quad-\quad \frac { 1 }{ 8 } +\frac { 1 }{ 8 } \cos\ { 2 } { A } \right]\\=\frac { 2 }{ 8 } \sin\ { A } +\frac { 4 }{ 8 } \sin\ { A } .\cos\ { 2 } { A } \\ but, \\ \sin\ { A } .\cos\ { 2 } { A }=\frac { 1 }{ 2 } \left[ \sin\ { 3A } -\sin\ { A } \right] \\ then, \\ =\frac { 2 }{ 8 } \sin\ { A } +\frac { 2 }{ 8 } \sin\ { 3A } -\frac { 2 }{ 8 } \sin\ { A } =\frac { 1 }{ 4 } \sin\ { A } \\ So,\quad the\quad answer\quad is\quad x\quad =\quad 4 \\ 1.5x\quad =\quad \boxed { 6 } \end{matrix}$

Since 4sin(60-A)sinAsin(60+A)=sin3A So 1.5*4=6.