Question

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

• A. \(2|\sin x|=\sin y\)
• B. \(2\sin x=\sin y\)
• C. \(\sin x=2\sin y\)
• D. \(\sin x=|\sin y|\)

Hint:

When exponential inequalities involve bounded trigonometric functions, first determine the minimum and maximum possible values of each term.

Answer 1

The Correct Option is D

Step 1: Simplify the given inequality.
Given, \[ 2^{\sqrt{\sin^2 x-2\sin x+5}}-\frac{1}{4^{\sin y}}\leq 1 \] Rewrite the second term: \[ \frac{1}{4^{\sin y}} = 4^{-\sin y} = 2^{-2\sin y} \] Hence, \[ 2^{\sqrt{\sin^2 x-2\sin x+5}}-2^{-2\sin y}\leq 1 \]

Step 2: Simplify the square root expression.
Observe that \[ \sin^2 x-2\sin x+5 = (\sin x-1)^2+4 \] Since \[ (\sin x-1)^2\geq 0, \] we get \[ (\sin x-1)^2+4\geq 4 \] Thus, \[ \sqrt{\sin^2 x-2\sin x+5}\geq 2 \] Therefore, \[ 2^{\sqrt{\sin^2 x-2\sin x+5}}\geq 2^2=4 \] Also, \[ -1\leq \sin y\leq 1 \] Hence, \[ 2^{-2\sin y}\leq 4 \] Therefore, \[ 2^{\sqrt{\sin^2 x-2\sin x+5}}-2^{-2\sin y}\geq 0 \] For the inequality to satisfy \[ 2^{\sqrt{\sin^2 x-2\sin x+5}}-2^{-2\sin y}\leq 1, \] the extreme values must occur simultaneously.

Step 3: Equality condition.
We must have \[ 2^{\sqrt{\sin^2 x-2\sin x+5}}=4 \] So, \[ \sqrt{\sin^2 x-2\sin x+5}=2 \] Squaring, \[ \sin^2 x-2\sin x+5=4 \] \[ \sin^2 x-2\sin x+1=0 \] \[ (\sin x-1)^2=0 \] Thus, \[ \sin x=1 \] Also, \[ 2^{-2\sin y}=4 \] \[ -2\sin y=2 \] \[ \sin y=-1 \] Hence, \[ \sin x=1 \quad \text{and} \quad \sin y=-1 \] Therefore, \[ \sin x=|\sin y| \]

Step 4: Final conclusion.
Hence, all such pairs satisfy \[ \boxed{\sin x=|\sin y|} \]