This question was uploaded on 14/11/21 on social media accounts.
This question was asked in IIT JEE Mains exam.
Solutions:
From 1st equation:
\[3\sin^2x=1-2\sin^2y\]
\[\Rightarrow 3\sin^2x=\cos{2y}\]
From 2nd equation:
\[3\sin{2x}=2\sin{2y}\]
\[\Rightarrow 3\sin x \cos x =2\sin{2y}\]
By dividing the above equations:
\[\frac{3\sin^2x}{3\sin x \cos x} =\frac{2\cos{2y}}{2\sin{2y}}\]
\[\Rightarrow \tan x =\cot{2y}\]
\[\Rightarrow x+2y = \frac\pi2 + \color{blue}{2n\pi}\]
