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Informaţii, definiţii, teoreme, formule, exerciţii şi probleme rezolvate din matematica de liceu.

Data publicarii: 17 August, 2016

EXERCITIUL 13

Suport teoretic:

Determinanti,proprietatile determinantilor,functii trigonometrice,identitati trigonometrice,ecuatii trigonometrice.

Enunt:

Fie determinantul:

\Delta(x)=\begin{vmatrix}1&1&1\\sinx&sin2x&sin3x\\cosx&cos2x&cos3x\end{vmatrix}\;.\Delta(x)=\begin{vmatrix}1&1&1\\sinx&sin2x&sin3x\\cosx&cos2x&cos3x\end{vmatrix}\;.  

Sa se rezolve ecuatia Δ(x) = sin2x, xЄR .

Raspuns: 

\;x\in{\{k\pi}\}\cup{\{\pm{arccos(-\frac{1}{3})+2k\pi}\}\;,k\in{\mathbb{Z}}}\;.\;x\in{\{k\pi}\}\cup{\{\pm{arccos(-\frac{1}{3})+2k\pi}\}\;,k\in{\mathbb{Z}}}\;.  

Rezolvare:

Aplicand proprietati uzuale ale determinantilor, avem, succesiv :

\Delta(x)=\begin{vmatrix}1&1&0\\sinx&sin2x&(sin3x-sin2x)\\cosx&cos2x&(cos3x-cos2x)\end{vmatrix}=\begin{vmatrix}1&0&0\\sinx&(sin2x-sinx)&(sin3x-sin2x)\\cosx&(cos2x-cosx)&(cos3x-cos2x)\end{vmatrix}=\Delta(x)=\begin{vmatrix}1&1&0\\sinx&sin2x&(sin3x-sin2x)\\cosx&cos2x&(cos3x-cos2x)\end{vmatrix}=\begin{vmatrix}1&0&0\\sinx&(sin2x-sinx)&(sin3x-sin2x)\\cosx&(cos2x-cosx)&(cos3x-cos2x)\end{vmatrix}=

=\begin{vmatrix}(sin2x-sinx)&(sin3x-sin2x)\\(cos2x-cosx)&(cos3x-cos2x)\end{vmatrix}=\begin{vmatrix}2sin{\frac{x}{2}}cos{\frac{3x}{2}}&2sin{\frac{x}{2}}cos{\frac{5x}{2}}\\-2sin{\frac{3x}{2}}sin{\frac{x}{2}}&-2sin{\frac{5x}{2}}sin{\frac{x}{2}}\end{vmatrix}==\begin{vmatrix}(sin2x-sinx)&(sin3x-sin2x)\\(cos2x-cosx)&(cos3x-cos2x)\end{vmatrix}=\begin{vmatrix}2sin{\frac{x}{2}}cos{\frac{3x}{2}}&2sin{\frac{x}{2}}cos{\frac{5x}{2}}\\-2sin{\frac{3x}{2}}sin{\frac{x}{2}}&-2sin{\frac{5x}{2}}sin{\frac{x}{2}}\end{vmatrix}=

=-{sin^2\frac{x}{2}}\cdot{\begin{vmatrix}cos\frac{3x}{2}&sin\frac{5x}{2}\\sin\frac{3x}{2}&sin\frac{5x}{2}\end{vmatrix}}={-(sin^2\frac{x}{2})}\cdot{(sin\frac{5x}{2}cos\frac{3x}{2}-sin\frac{3x}{2}cos\frac{5x}{2})}==-{sin^2\frac{x}{2}}\cdot{\begin{vmatrix}cos\frac{3x}{2}&sin\frac{5x}{2}\\sin\frac{3x}{2}&sin\frac{5x}{2}\end{vmatrix}}={-(sin^2\frac{x}{2})}\cdot{(sin\frac{5x}{2}cos\frac{3x}{2}-sin\frac{3x}{2}cos\frac{5x}{2})}=

=-{sin^2\frac{x}{2}}\cdot{sin(\frac{5x}{2}-\frac{3x}{2}}=-{sin^2\frac{x}{2}}\cdot{sinx}\;.=-{sin^2\frac{x}{2}}\cdot{sin(\frac{5x}{2}-\frac{3x}{2}}=-{sin^2\frac{x}{2}}\cdot{sinx}\;.  

Ecuatia  Δ(x) = sin2x devine, succesiv:

-{sin^2\frac{x}{2}}\cdot{sinx}=sin2x\Leftrightarrow-{sin^2\frac{x}{2}}\cdot{sinx}=2sinxcosx\Leftrightarrow-{sin^2\frac{x}{2}}\cdot{sinx}=sin2x\Leftrightarrow-{sin^2\frac{x}{2}}\cdot{sinx}=2sinxcosx\Leftrightarrow

\Leftrightarrow(sinx=0)\;sau\;(-{sin^2\frac{x}{2}}=2cosx)\;\Leftrightarrow(sinx=0)\;sau\;(-{sin^2\frac{x}{2}}=2cosx)\;

1)\;{sinx=0}\Leftrightarrow{x=k\pi,\;unde\;k\in{\mathbb{Z}}}\;.1)\;{sinx=0}\Leftrightarrow{x=k\pi,\;unde\;k\in{\mathbb{Z}}}\;.

2)\;{-{sin^2\frac{x}{2}}=2cosx}\Leftrightarrow{-\frac{1-cosx}{2}=2cosx}\Leftrightarrow{cosx=-\frac{1}{3}}\Leftrightarrow2)\;{-{sin^2\frac{x}{2}}=2cosx}\Leftrightarrow{-\frac{1-cosx}{2}=2cosx}\Leftrightarrow{cosx=-\frac{1}{3}}\Leftrightarrow

\Leftrightarrow{x=\pm{arccos(-\frac{1}{3})+2k\pi}\;,k\in{\mathbb{Z}}}\;.\Leftrightarrow{x=\pm{arccos(-\frac{1}{3})+2k\pi}\;,k\in{\mathbb{Z}}}\;.

Deci: 

\;x\in{\{k\pi}\}\cup{\{\pm{arccos(-\frac{1}{3})+2k\pi}\}\;,k\in{\mathbb{Z}}}\;.\;x\in{\{k\pi}\}\cup{\{\pm{arccos(-\frac{1}{3})+2k\pi}\}\;,k\in{\mathbb{Z}}}\;.

Postat în: DETERMINANTI-liceu

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