Find the number of points in the region $x^{2}+y^{2}\leq4$ satisfying the equation $\tan^4x+\cot^4x+1=3\sin^2y$.

Question

Given,

$x^{2}+y^{2}\leq4$

$\tan^4x+\cot^4x+1=3\sin^2y$.

It's a past problem of an UG entrance.I tried it solving using the graphical method,but couldn't.And also using trigonometric transformation I didn't get anything.I think I'm lacking some conceptual information.Please help me to get it out.

Answer 1

Note that $\tan x \times \cot x =1$ hence $\tan^4 x + \cot ^4x$ takes a minimum of $2$ when $x=\pm \pi/4$, the only values for $x$ in your disk.

Can you continue from here?

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Do we agree on the followings?

1. The right hand side is $3 \sin^2 y$, which can have a maximum of $3$.
2. The left hand side is $1+\tan^4 x + \cot ^4x$ which can take a min of $3$.

If we agree then we can see that the equality can hold only when the right side is at its max and the left side is at its min.

Answer 2

I see that the other answer posted here is much simpler, but here is what I came up with: Using basic trig identities, $tan^4x + cot^4x + 1 = 3sin^2y$ becomes $(sec^2x - 1)^2 + (csc^2x - 1)^2 + 1 = 3sin^2y$. Now for some manipulation: $$sec^4x - 2sec^2x + 1 + csc^4x - 2csc^2x + 1 + 1 = 3sin^2y$$ $$ \rightarrow sec^4x + csc^4x - 2(sec^2x + csc^2x) + 3 = 3sin^2y$$ $$ \rightarrow sec^4x + csc^4x - 2(sec^2x + csc^2x) = 3sin^2y - 3$$ $$ \rightarrow \frac{sin^4x + cos^4x}{sin^4xcos^4x} - \frac{2}{sin^2xcos^2x} = -3cos^2y$$ $$ \rightarrow \frac{sin^4x - 2sin^2xcos^2x + cos^4x}{sin^4xcos^4x} = -3cos^2y$$ $$\rightarrow 16\frac{(sin^2x - cos^2x)^2}{sin^4(2x)} = -3cos^2y$$ $$ \rightarrow 16\frac{cos^2(2x)}{sin^4(2x)} = -3cos^2y$$ $$\rightarrow 16cot^2(2x)csc^2(2x) = -3cos^2y$$

Since the expression on the left must be positive or zero and the expression on the right must be negative or zero, the only way they can be equal is if both are zero. This should allow you to find all values of $x$ and $y$ meeting the conditions.