Sa se rezolve ecuatia: \( \sin x\cdot \sin y\cdot \sin z+\cos x\cdot \cos y\cdot \cos z=1 \).
Concursul Victor Valcovici, 2001
Ecuatie trigonometrica
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Mateescu Constantin
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Deoarece \( 1=|\sin x\sin y\sin z+\cos x\cos y\cos z|\le |\sin x\sin y|+|\cos x\cos y|=\pm \cos(x\pm y)\le1 \) si analoagele rezulta ca fiecare sinus si cosinus trebuie sa fie 0 , 1 sau -1.
Deasemenea cel putin unul trebuie sa fie 0.
De exemplu, daca \( \sin x=0 \) ,\( \cos x=\cos y=\cos z=1 \) adica \( x,y,z\in\{2k\pi\} , k\in\mathbb{Z} \)
sau \( x,y\in\{(2k+1)\pi\}, z\in\{2k\pi\},k\in\mathbb{Z} \) si permutari circulare.
Daca \( \cos x=0 \) atunci \( x,y,z\in\{\frac{(4k+1)\pi}{2}\},k\in\mathbb{Z} \) sau \( x,y\in\{\frac{(4k+3)\pi}{2}\},z\in\{\frac{(4k+1)\pi}{2}\},k\in\mathbb{Z} \) si permutari circulare.
Deasemenea cel putin unul trebuie sa fie 0.
De exemplu, daca \( \sin x=0 \) ,\( \cos x=\cos y=\cos z=1 \) adica \( x,y,z\in\{2k\pi\} , k\in\mathbb{Z} \)
sau \( x,y\in\{(2k+1)\pi\}, z\in\{2k\pi\},k\in\mathbb{Z} \) si permutari circulare.
Daca \( \cos x=0 \) atunci \( x,y,z\in\{\frac{(4k+1)\pi}{2}\},k\in\mathbb{Z} \) sau \( x,y\in\{\frac{(4k+3)\pi}{2}\},z\in\{\frac{(4k+1)\pi}{2}\},k\in\mathbb{Z} \) si permutari circulare.