All the pairs (x,y) that satisfy the inequality
\[2^{\sqrt{\sin^2x-2\sin x+5}} \cdot \frac{1}{4^{\sin^2 y}} \le 1\]
also satisfy the equation:
Solutions · 1
Official solution
\[2^{\sqrt{\sin^2x-2\sin x+5}} \le 4^{\sin^2y} = 2^{2\sin^2y} \implies \sqrt{\sin^2x-2\sin x+5} \le 2\sin^2y\]
\[\implies \sqrt{(\sin x-1)^2+4} \le 2\sin^2y\]
The left side has minimum value 0+4=2, attained at sinx=1. The right side has maximum value 2, attained at sin2y=1. So the inequality can only hold with equality on both sides simultaneously:
\[\sin x = 1 \quad \text{and} \quad |\sin y|=1\]
Therefore sinx=∣siny∣.