All the pairs (x, y) that satisfy the inequality \({2^{\sqrt {{\rm{si}}{{\rm{n}}^2}{\rm{x}} - 2{\rm{sinx}} + 5} }}\cdot\frac{1}{{{4^{{\rm{si}}{{\rm{n}}^2}{\rm{y}}}}}} \le 1\) also satisfy the equation:

All the pairs (x, y) that satisfy the inequality \({2^{\sqrt {{\rm{si}

1.	All the pairs (x, y) that satisfy the inequality \({2^{\sqrt {{\rm{si}}{{\rm{n}}^2}{\rm{x}} - 2{\rm{sinx}} + 5} }}\cdot\frac{1}{{{4^{{\rm{si}}{{\rm{n}}^2}{\rm{y}}}}}} \le 1\) also satisfy the equation:
A.	2\|sin x\| = 3sin y
B.	2sin x = sin y
C.	sin x = 2sin y
D.	sin x = \|sin y\|
Answer» E.

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All the pairs (x, y) that satisfy the inequality \({2^{\sqrt {{\rm{si}}{{\rm{n}}^2}{\rm{x}} - 2{\rm{sinx}} + 5} }}\cdot\frac{1}{{{4^{{\rm{si}}{{\rm{n}}^2}{\rm{y}}}}}} \le 1\) also satisfy the equation:
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All the pairs (x, y) that satisfy the inequality \({2^{\sqrt {{\rm{si}}{{\rm{n}}^2}{\rm{x}} - 2{\rm{sinx}} + 5} }}\cdot\frac{1}{{{4^{{\rm{si}}{{\rm{n}}^2}{\rm{y}}}}}} \le 1\) also satisfy the equation:

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