please finish trigonometric equation.

Oct 2009
7
0
Can you show me where this goes, or start me on the right path?

\(\displaystyle 3cot^3x - cotx = \sqrt {3}\)

\(\displaystyle \frac {3cos^3x}{ sin^3x}-\frac{cosx}{sinx}= \sqrt {3}\)

\(\displaystyle \frac {3cos^3x}{ sin^3x}-\frac{cosx\times sin^2x}{sin^3x}=\sqrt {3}\)

\(\displaystyle \frac {3cos^3x-cosx\times sin^2x}{ sin^3x}=\sqrt {3}\)