Ecuatie trigonometrica

[Al3xx]

Sa se determine \( x,y \in R \) stiind ca:


\( tg^4x + tg^4y +2ctg^2x ctg^2y = 3+sin^2(x+y) \)

[Virgil Nicula]

\( \tan^4x+\tan^4y+\frac {2}{\tan^2x\tan^2y}=\left(\tan^2x-\tan^2y\right)^2+2\left(\tan^2x\tan^2y+\frac {1}{\tan^2x\tan^2y}\right)\ge 4\ge 3+\sin^2(x+y) \) .

Deci solutiile ecuatiei date sunt aceleasi cu solutiile sistemului de ecuatii \( \left\|\begin{array}{c}
\tan^2x=\tan^2y=1\\\\\\\\
\sin^2(x+y)=1\end{array}\ \ \) a.s.o.