Mock Olympiad Math Problem: Solve $\large 3^{\sin^4 x-\cos^2 x}-3^{\cos^4 x-\sin^2 x}=\cos 2x$.

Solve $\large 3^{\sin^4 x-\cos^2 x}-3^{\cos^4 x-\sin^2 x}=\cos 2x$.

$\large 3^{\sin^4 x-\cos^2 x}-3^{\cos^4 x-\sin^2 x}=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}\left(1-\dfrac{3^{\cos^4 x-\sin^2 x}}{3^{\sin^4 x-\cos^2 x}}\right)=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}(1-3^{(\cos^4 x-\sin^2 x)-(\sin^4 x-\cos^2 x)})=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}(1-3^{(\cos^4 x-\sin^2 x-\sin^4 x+\cos^2 x)})=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}(1-3^{(\cos^4 x-\sin^2 x-\sin^4 x+\cos^2 x)})=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}(1-3^{(\cos^2 x)^2-\sin^2 x-\sin^4 x+\cos^2 x)})=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}(1-3^{(1-\sin^2 x)^2-\sin^2 x-\sin^4 x+1-\sin^2 x)})=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}(1-3^{1-2\sin^2 x+\sin^4 x-\sin^2 x-\sin^4 x+1-\sin^2 x)})=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}(1-3^{2-4\sin^2 x})=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}(1-3^{2(1-2\sin^2 x)})=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}(1-3^{2\cos 2x})=\cos 2x$

$\large 3^{\sin^4 x-\cos^2 x}(1-3^{2\cos 2x})(1-3^{2\cos 2x})=\cos 2x(1-3^{2\cos 2x})$

$\large 3^{\sin^4 x-\cos^2 x}(1-3^{2\cos 2x})^2=\cos 2x(1-3^{2\cos 2x})$

$\large 2(3^{\sin^4 x-\cos^2 x})(1-3^{2\cos 2x})^2=2\cos 2x(1-3^{2\cos 2x})$---(*)

Since as we have already explained here (recruit-for-attitude-train-for-skills), we know full well that:

$y=f(x)(1-3^{f(x)})\le 0$ for all $x$

Applying this to the equation (*), we noticed that:

1.  we have a squared term on the LHS that is always greater than or equal to zero: $(1-3^{2\cos 2x})^2\ge 0$,

2.  we have the $ 2(3^{\sin^4 x-\cos^2 x})$ that is always greater than zero.

Putting all these observations together we get:

$\text{always positive}\cdot{\text{always greater than or equal to zero }}=\text{always less than or equal to zero }$

Here comes the most crucial part in solving the problem effectively and efficiently! The above observation requires us to think logically, what does it mean exactly by something that is always positive, when it is multiplied by another quantity that could be zero or positive, will always yield a negative or a zero?

Ah!!! This can be true iff both sides equal to zero, and that happens when:

$\cos 2x=0$

Therefore, $2x=\dfrac{\pi}{2}+n\pi$ for integer $n$.

That is, $x=\dfrac{\pi}{4}+\dfrac{n\pi}{2}$ for integer $n$.