<s>[INST] Find all $(x,y)$ satisfying $(\sin^2x+\frac{1}{\sin^2 x})^2+(\cos^2x+\frac{1}{\cos^2 x})^2=12+\frac{1}{2}\sin y$ Find all pairs $(x,y)$ of real numbers that satisfy the equation $(\sin^2x+\frac{1}{\sin^2 x})^2+(\cos^2x+\frac{1}{\cos^2 x})^2=12+\frac{1}{2}\sin y$

I supposed $a=\sin^2x$ and $b=\cos^2x$
So the equation becomes $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2=12+\frac{1}{2}\sin y$
As $a+\frac{1}{a}\geq 2$ and $b+\frac{1}{b}\geq 2$
$12+\frac{1}{2}\sin y\geq 8$
$\sin y\geq -8$
I am stuck here.I could not solve further.Please help me.Thanks.
 [/INST] Expanding $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 = 12 + \sin y$
will get us $a^2 + 2 + \frac{1}{a^2} + b^2 + 2 + \frac{1}{b^2} = 12 + \sin y$.
And we can subtract 4 from both sides.
$a^2 + b^2 + \frac{1}{a^2} + \frac{1}{b^2} = 8 + \frac{1}{2}\sin y$
Now make y as a function of x:
$y = asin(2a^2 + 2b^2 + \frac{2}{a^2} + \frac{2}{b^2} - 16)$
Substitute back a and b to get
$y = asin(2sin^4x + 2cos^4x + 2sec^4x + 2csc^4x - 16)$
Because $asin(x)$ is only defined for $0 \leq x \leq 1$, we can say that
$0 \leq 2sin^4x + 2cos^4x + 2sec^4x + 2csc^4x - 16 \leq 1$ or $16 \leq 2sin^4x + 2cos^4x + 2sec^4x + 2csc^4x \leq 17$.
Now we can complete the square with $2sin^4x + 2cos^4x$ to get $16 \leq 2(sin^2x + cos^2x)^2 - 4sin^2xcos^2x + 2sec^4x + 2csc^4x \leq 17$.
This reduces to $16 \leq 2 - 4sin^2xcos^2x + 2sec^4x + 2csc^4x \leq 17$ which can be rewritten as $14 \leq 2sec^4x + 2csc^4x - 4sin^2xcos^2x \leq 15$.
Using some trigonometric identities we get $14 \leq 2sec^4x + 2csc^4x - sin^2(2x) \leq 15$.
So we have y as a function of x, we have the domain, and we have the range.
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