All the pairs (x, y) that satisfy the inequality
$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$$
also satisfy the equation

If $$\alpha $$ and $$\beta $$ are the roots of the quadratic equation,
x² + x sin $$\theta $$ - 2 sin $$\theta $$ = 0, $$\theta \in \left( {0,{\pi \over 2}} \right)$$, then
$${{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}$$ is equal to :

If m is chosen in the quadratic equation

(m² + 1) x² – 3x + (m² + 1)² = 0

such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :-

Let p, q $$ \in $$ R. If 2 - $$\sqrt 3$$ is a root of the quadratic equation, x² + px + q = 0, then :