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Ecuatie trigonometrica.
Posted: Sun Jun 06, 2010 2:24 am
by Virgil Nicula
Sa se rezolve ecuatia \( \sin\ \left(\ 45^{\circ}\ +\ x\ \right)\ \cdot\ \sin\ 15^{\circ}=\sin\ x\ \cdot\ \sin\ 30^{\circ} \) , unde \( x\ \in\ \left(\ 0\ ,\ \frac {\pi}{2}\ \right) \) .
Posted: Sun Jun 06, 2010 11:21 am
by Mateescu Constantin
Ecuatia se scrie : \( (\sin\ x\ \cdot\ \cos\ 45^{\circ}\ +\ \cos\ x\ \cdot\ \sin\ 45^{\circ})\ \cdot\ \sqrt{\frac{1-\cos 30^{\circ}}{2}}\ =\ \sin\ x\ \cdot\ \frac 12 \)
\( \Longleftrightarrow\ \frac{\sqrt 2}{2}\ \cdot\ (\sin\ x\ +\ \cos\ x)\ \cdot\ \frac{\sqrt{2-\sqrt 3}}{2}\ =\ \sin\ x\ \cdot\ \frac 12\ \Longleftrightarrow\ \sqrt 2\ \cdot\ (\sin\ x\ +\ \cos\ x)\ \cdot\ \frac{\sqrt 6-\sqrt 2}{4}\ =\ \sin\ x \)
\( \Longleftrightarrow\ (\sin\ x\ +\ \cos\ x)\ \cdot\ \frac{\sqrt 3-1}{2}\ =\ \sin\ x \) . Cum \( x\ \in\ \left(\ 0\ ,\ \frac{\pi}2\ \right) \) putem imparti prin \( \ \sin\ x \) si obtinem :
\( 1\ +\ \cot\ x\ =\ \frac{2}{\sqrt 3-1}\ \Longleftrightarrow\ \cot\ x\ =\ \sqrt 3 \) , deci \( x\ =\ 30^{\circ} \) este solutia ecuatiei .