KVPY SX SB Exam 05-Nov-2017 Question Paper

only possible when sin 3x = 1 & sin 7x = 1
sin 3x = 1
$\sin 3\mathit{x}=\sin \left(4\mathit{n}+1\right)\frac{\mathit{\pi }}{2},\mathit{n}\in \mathit{I}$
$3\mathit{x}=\left(4\mathit{n}+1\right)\frac{\mathit{\pi }}{2}\Rightarrow \mathit{x}=\left(4\mathit{n}+1\right)\frac{\mathit{\pi }}{6}$
$\sin 7\mathit{x}=\sin \left(4\mathit{m}+1\right)\frac{\mathit{\pi }}{2},\mathit{m}\in \mathit{I}$
$\mathit{x}=\left(4\mathit{m}+1\right)\frac{\mathit{\pi }}{14}$
for common solution
$\left(4\mathit{n}+1\right)\frac{\mathit{\pi }}{6}=\left(4\mathit{m}+1\right)\frac{\mathit{\pi }}{14}$
Solving these 1 = 3m – 7n
First solution is m = 5, n = 2
Second solution is m = 12, n = 5
So two solutions are possible