Animated Solution for Mathematics - Trigonometry: All the pairs $(x,y)$ that satisfy the inequality $2^{{\sin }^{2}x}\cdot 2^{\sin x}\cdot 5\cdot 4^{-{\sin }^{2}y}\le 1$ also satisfy the equation.

The given inequality is:

To simplify this, we must unify the bases. Since $4=2^2$, we can rewrite the second term as:

Substituting this back into the original inequality, we get:

Using the laws of exponents, we combine the terms on the left side:

Since the base $2>1$, the exponential function is strictly increasing. We can therefore compare the exponents directly:

Rearranging the terms to isolate the variables, we obtain:

Focusing on the expression ${\sin }^{2}x-2\sin x+3$, we complete the square by splitting the constant $3$ into $1+2$:

We know that for any real $x$, $(\sin x-1{)}^2\ge 0$. Thus, the minimum value of the left-hand side is $0+2=2$.

We also know that the range of ${\sin }^{2}y$ is $[0,1]$. Consequently, the range of the right-hand side is:

For the inequality $(\sin x-1{)}^2+2\le 2{\sin }^{2}y$ to hold, the left side (which is at least $2$) must be less than or equal to the right side (which is at most $2$). This is only possible if both sides are exactly equal to $2$.

This leads to the following conditions:

Thus, the solution is defined by the condition $\sin x=|\sin y|$.