Find all $(x,y)$ satisfying $(\sin^2x+\frac{1}{\sin^2 x})^2+(\cos^2x+\frac{1}{\cos^2 x})^2=12+\frac{1}{2}\sin y$

Question

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

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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.

Answer 1

As JMoravitz notes in the comments, it's probably the case that the minimum value of the LHS is 12.5. Let's prove this.

Using the inequality $a^2+b^2\ge\frac{(a+b)^2}{2}$, note that \begin{align*}\left(\sin^2x + \frac{1}{\sin^2x}\right)^2 + \left(\cos^2x + \frac{1}{\cos^2x}\right)^2&\ge \frac{1}{2}\left(\sin^2x + \frac{1}{\sin^2x} + \cos^2x + \frac{1}{\cos^2x}\right)^2\\ &=\frac{1}{2}\left(1 + \frac{\cos^2x+\sin^2x}{\sin^2x\cos^2x}\right)^2 \\ &=\frac{1}{2}\left(1 + \frac{4}{(2\sin x\cos x)^2}\right)^2 \\ &=\frac{1}{2}\left(1 + \frac{4}{\sin^2{2x}}\right)^2\\ &\ge\frac{1}{2}\left(1 + \frac{4}{1}\right)^2 = \frac{25}{2}. \end{align*} I leave it to you to determine when this equality holds.

Answer 2

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.