Problema Functii trigonometrice

Isac Dum Feb 12, 2012 11:25 pm

Ma juati sa rezolv aceste 2 exercitii?
1) rad3*sin2x+cos2x=rad2
2) 2sin(x/2)+rad3cos(x/2)=2
Trebuie sa rezolv aceste exercitiu prin metoda substitutiei universale si metoda omogenitatii!!!Ma ajutati am test maine!!!Si vreau sa am aceste 2 exercitii rezolvate ,consider eu ca sunt mai dificile decat celelate si cred ca o sa imi cada unul asemanator!!!

**Admin** Lun Feb 13, 2012 8:36 pm

Isac a scris:1) rad3*sin2x+cos2x=rad2

Folosind metoda omogenități, putem scrie :
$\begin{array}{l} \sqrt 3 \times \sin \left( {2x} \right) + \cos \left( {2x} \right) = \sqrt 2 \\ \left\{ \begin{array}{l} \sqrt 3 \sin \left( {2x} \right) = 2\sqrt 3 \sin x\cos x\\ \cos \left( {2x} \right) = {\cos ^2}x - {\sin ^2}x \end{array} \right. \Rightarrow \\ 2\sqrt 3 \sin x\cos x + {\cos ^2}x - {\sin ^2}x = \sqrt 2 \left( {{{\cos }^2}x + {{\sin }^2}x} \right)\left| {\left( { \div {{\sin }^2}x} \right)} \right.\\ 2\sqrt 3 \frac{{\cos x}}{{\sin x}} + {\left( {\frac{{\cos x}}{{\sin x}}} \right)^2} - 1 = \sqrt 2 {\left( {\frac{{\cos x}}{{\sin x}}} \right)^2} + \sqrt 2 \\ t = \frac{{\cos x}}{{\sin x}} \Rightarrow 2\sqrt 3 t + {t^2} - 1 = \sqrt 2 {t^2} + \sqrt 2 \Leftrightarrow {t^2}\left( {1 - \sqrt 2 } \right) + 2\sqrt 3 t - 1 - \sqrt 2 = 0\\ t = - 2 - \sqrt 2 + \sqrt 3 + \sqrt 6 \Rightarrow ctgx = - 2 - \sqrt 2 + \sqrt 3 + \sqrt 6 \Rightarrow x = arcctg\left( { - 2 - \sqrt 2 + \sqrt 3 + \sqrt 6 } \right) + k\pi |k \in Z \end{array}$

Problema Functii trigonometrice

Problema Functii trigonometrice

Re: Problema Functii trigonometrice