MHB Solving for $(x,y)$ in $\sin(2x)$

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The discussion focuses on finding pairs of real numbers $(x, y)$ that satisfy the equation involving $\sin(2x)$ for $0 < x < \frac{\pi}{2}$. By applying the AM-GM inequality, it is shown that the left-hand side of the equation can be bounded, leading to the conclusion that $y - \frac{y^2}{4} \ge 1$. This inequality simplifies to $(y - 2)^2 \le 0$, indicating that the only solution is $y = 2$. The unique solution for the given conditions is $(x, y) = \left(\frac{\pi}{4}, 2\right)$.
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Find all pairs $(x, y)$ of real numbers with $0<x<\dfrac{\pi}{2}$ such that $$\frac{(\sin x)^{2y}}{(\cos x)^{\frac{y^2}{2}}}+\frac{(\cos x)^{2y}}{(\sin x)^{\frac{y^2}{2}}}=\sin (2x)$$.
 
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Here is a solution that I found somewhere online:

By applying AM-GM inequality to the expression on the left, we obtain

$$\frac{(\sin x)^{2y}}{(\cos x)^{\frac{y^2}{2}}}+\frac{(\cos x)^{2y}}{(\sin x)^{\frac{y^2}{2}}} \ge 2 \sqrt{\frac{(\sin x \cos x)^{2y}}{(\sin x\cos x)^{\frac{y^2}{2}}}} \ge 2(\sin x \cos x)^{y-\frac{y^2}{4}} $$

and since $$\frac{(\sin x)^{2y}}{(\cos x)^{\frac{y^2}{2}}}+\frac{(\cos x)^{2y}}{(\sin x)^{\frac{y^2}{2}}}=\sin (2x)$$

$$\sin (2x) \ge 2(\sin x \cos x)^{y-\frac{y^2}{4}}$$

$$2(\sin x \cos x) \ge 2(\sin x \cos x)^{y-\frac{y^2}{4}}$$

but $$\sin x \cos x=\frac{\sin 2x}{2} \le \frac{1}{2}$$

we must have $$y-\frac{y^2}{4}\ge 1\rightarrow (y-2)^2 \le 0$$

This is true iff $y=2$ and when $\sin x=\cos x$, so there is a unique solution to this problem where $$(x, y)=\left(\frac{\pi}{4}, 2 \right)$$.
 
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