Solving the trigonometric equation $\tan^2x+\cot^2x=2-\cos^{2014}(2x)$

Question

I was solving the trigonometric equation $$\tan^2x+\cot^2x=2-\cos^{2014}(2x) $$ I solve it by inequality $|a|+\frac{1}{|a| }\geq 2$.

$$ L.H.S=\tan^2x+\cot^2x =\tan^2x+\frac{1}{\tan^2x} \xrightarrow[|a|+\frac{1}{|a| }\geq 2]{}\geq 2 $$ $$R.H.S=2-\cos^{2014}(2x) \xrightarrow[\cos^{2n}(2x) \geq 0]{}2-\cos^{2014}(2x) \leq 2$$ so $$R.H.S \leq 2\\ L.H.S \geq2 \rightarrow R.H.S = L.H.S=2 \\\tan^2x=1 \rightarrow x=\pm \frac{\pi}{4}+k\pi \\\cos^{2014}(2x)=0 \rightarrow 2x=\pm \frac{\pi}{2} +2k\pi \rightarrow x=\pm \frac{\pi}{4}+k\pi $$

Can someone give an alternative proof?

Answer 1

The inequality proof looks fine, but if you need another way, let $\cos x =t$, so $|t|<1$. The equation is then $$\frac{1-t^2}{t^2}+\frac{t^2}{1-t^2}-2+(2t^2-1)^{2014}=0$$ $$\iff \frac{(2t^2-1)^2}{(1-t^2)t^2}+(2t^2-1)^{2014}=0$$

As both terms are non-negative, we must have $2t^2-1=0\implies \cos^2x=\frac12$.