All the pairs (x, y) that satisfy the inequality [2^ ^2x - 2 x + 5 .1 4^ ^2y 1] also satis…

Question

All the pairs (x, y) that satisfy the inequality

$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$$

also satisfy the equation

Accepted Answer

$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }} \le {2^{2{{\sin }^2}y}}$$

$$ \Rightarrow $$ $$\sqrt {{{\sin }^2}x - 2\sin x + 5}  \le 2{\sin ^2}y$$

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

it is true when sinx = 1, |siny| = 1

so sinx = |siny|

Solutions · 1