In this triangle, we have
[tex]\cos30^\circ=\dfrac{4x\sqrt3-5\sqrt3}{5x-4}[/tex]
You should know that [tex]\cos30^\circ=\dfrac{\sqrt3}2[/tex], so that
[tex]\dfrac{\sqrt3}2=\dfrac{4x\sqrt3-5\sqrt3}{5x-4}[/tex]
Divide both sides by [tex]\sqrt3[/tex] to get
[tex]\dfrac12=\dfrac{4x-5}{5x-4}[/tex]
Multiply both sides by [tex]5x-4[/tex] to get
[tex]\dfrac{5x-4}2=4x-5[/tex]
Simplify and solve for [tex]x[/tex]:
[tex]\dfrac{5x}2-2=4x-5[/tex]
[tex]5-2=4x-\dfrac{5x}2[/tex]
[tex]3=\dfrac{3x}2[/tex]
[tex]6=3x[/tex]
[tex]\implies\boxed{x=2}[/tex]
