Question

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

• A. $\sin x=2\sin y$
• B. $2\sin x=\sin y$
• C. $\sin x=\left|\sin y\right|$
• D. $2\left|\sin x\right|=3\sin y$

Answer 1

The correct option is C $\sin x=\left|\sin y\right|$

$2^{\sqrt{{\sin }^{2}x-2\sin x+5}}\cdot \frac{1}{4^{{\sin }^{2}y}}\le 1$
$\Rightarrow 2^{\sqrt{{\sin }^{2}x-2\sin x+5}}\le 2^{2{\sin }^{2}y}\Rightarrow \sqrt{{\sin }^{2}x-2\sin x+5}\le 2{\sin }^{2}y\cdots \left(1\right)$


$\sqrt{{\sin }^{2}x-2\sin x+5}=\sqrt{(\sin x-1{)}^2+4}\Rightarrow \sqrt{(\sin x-1{)}^2+4}\in [2,2\sqrt{2}]$
But
$2{\sin }^{2}y\in [0,2]$
Equation $\left(1\right)$ holds true only when
$\sqrt{{\sin }^{2}x-2\sin x+5}=2{\sin }^{2}y=2$
$\Rightarrow (\sin x-1{)}^2+4=4\Rightarrow \sin x=1$
and $2{\sin }^{2}y=2\Rightarrow \left|\sin y\right|=1$
$\Rightarrow \sin x=\left|\sin y\right|$