2(sin^6 cos^6)-3(sin^4 cos^4) 1=0

Hi Sanjith,
Here is the answer to your question.
Given:
$\begin{array}{rl}& 2{\sin }^6\theta {\cos }^6\theta -3{\sin }^4\theta {\cos }^4\theta +1=0\\ & \text{Let}x={\sin }^2\theta {\cos }^2\theta \end{array}$
Then:
$2x^3-3x^2+1=0$
Factorising:
$(x-1)(2x^2-x-1)=0$
So, $x=1$ or $x=1$ or $x=-\frac{1}{2}$ .
Since $x={\sin }^2\theta {\cos }^2\theta \ge 0$ , valid value is $x=1$ .