If x and y are the solutions of the equation 12sin x + 5cos x = 2 y...

$LHS = 12sin x + 5cos x \in \left[- \sqrt{1{2}^{2} + {5}^{2}}, \sqrt{1{2}^{2} + {5}^{2}}\right]<br />i.e. \left[- 13,13\right] i.e.<br />maximum value$ of LHS$is$13$<br />$RHS = 2\left({y}^{2} - 4y + 4\right)$+ 13$
$= 2{\left(y - 2\right)}$^{2} + 13$<br />$RHS \ge 13$<br />Roots of the equation exist if$LHS = RHS = 13 .

$RHS = 13$ when $y = 2$
$LHS = 13 ⇒ 12sin x + 5cos x = 13$
$⇒ \dfrac{12}{13}sin x + \dfrac{5}{13}cos x = 1$
$\sin \left(x + \alpha\right) = 1,$ where $tan \alpha = \dfrac{5}{12}$
$x + ta{n}^{- 1}\dfrac{5}{12} = \dfrac{\pi}{2} ⇒ x = \dfrac{\pi}{2} - ta{n}^{- 1}\dfrac{5}{12}$
$⇒ xy = \pi - 2ta{n}^{- 1}\dfrac{5}{12} ⇒ \dfrac{5}{12} = \tan \left(\dfrac{\pi}{2} - \dfrac{xy}{2}\right) = \cot \left(\dfrac{xy}{2}\right)$
$⇒ 12\cot \left(\dfrac{xy}{2}\right) = 5$