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Usernameheureka
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Questions 11
Answers 3216

 #4
avatar+18566 
+5

...continued.

if x is in degrees:

$$\\cos(x) = cos\left[180\textcolor[rgb]{1,0,0}{\ensuremath{^\circ}}-cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]
\quad | \quad \pm cos^{-1}\\
\pm x=\pm \left[ 180\ensuremath{^\circ}} -cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]\\
x=180\ensuremath{^\circ}} -cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\\
cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]=180\ensuremath{^\circ}}-x
\quad | \quad \cos{}\\\\
\frac{0.25}{6,76}*\cos{(x)}=\cos{(180\ensuremath{^\circ}}-x)}\\
\frac{0.25}{6,76}*\cos{(x)}=\underbrace{\cos{180\ensuremath{^\circ}}}_{-1}\cos{(x)}
+\underbrace{\sin{180\ensuremath{^\circ}}}_{0}\sin{(x)}\\
\frac{0.25}{6,76}*\cos{(x)}=-\cos{(x)}\\
\frac{0.25}{6,76}*\cos{(x)}+\cos{(x)}=0\\
\cos{(x)}\left(\frac{0.25}{6,76}+1\right)=0\\
\cos{(x)}=0\quad | \quad \pm \cos^{-1}\\
\pm x=\frac{\pi}{2}\\\\
\boxed{x=\frac{\pi}{2} \pm 2\pi*k\qquad or \qquad x=-\frac{\pi}{2} \pm 2\pi*k}$$

\\cos(x) = cos\left[180\textcolor[rgb]{1,0,0}{\ensuremath{^\circ}}-cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]
\quad | \quad \pm cos^{-1}\\
\pm x=\pm \left[ 180\ensuremath{^\circ}} -cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]\\
x=180\ensuremath{^\circ}} -cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\\
cos^{-1}(\frac{0.25}{6,76}*\cos{(x)})\right]=180\ensuremath{^\circ}}-x
\quad | \quad \cos{}\\\\
\frac{0.25}{6,76}*\cos{(x)}=\cos{(180\ensuremath{^\circ}}-x)}\\
\frac{0.25}{6,76}*\cos{(x)}=\underbrace{\cos{180\ensuremath{^\circ}}}_{-1}\cos{(x)}
+\underbrace{\sin{180\ensuremath{^\circ}}}_{0}\sin{(x)}\\
\frac{0.25}{6,76}*\cos{(x)}=-\cos{(x)}\\
\frac{0.25}{6,76}*\cos{(x)}+\cos{(x)}=0\\
\cos{(x)}\left(\frac{0.25}{6,76}+1\right)=0\\
\cos{(x)}=0\quad | \quad  \pm \cos^{-1}\\
\pm x=\frac{\pi}{2}\\\\
\boxed{x=\frac{\pi}{2} \pm 2\pi*k\qquad or \qquad x=-\frac{\pi}{2} \pm 2\pi*k}




heureka May 16, 2014
 #8
avatar+18566 
+5

1.) See "Vieta"

2.) You find $$x_1=1$$ do:

$$\begin{array}{
rlllr
} (
x^3
&
+4x^2
&
+x
&
-6)
&
:
(x-1)
=\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6}
\\
\textcolor[rgb]{1,0,0}{{\underline{-(
x^3}}
&
\textcolor[rgb]{1,0,0}{\underline{-x^2)}}}
&
&
&
\\
0
&
+5x^2
&
+x
\\
&
\textcolor[rgb]{0,0,1}{\underline{-(5x^2}}
&
\textcolor[rgb]{0,0,1}{\underline{-5x)}}
\\
&
0
&
+6x
&
-6
\\
&&
\textcolor[rgb]{0,1,0}{\underline{-(6x}}
&
\textcolor[rgb]{0,1,0}{\underline{-6)}}
\\
&&
0
&
+0
\end{array}$$

\begin{array}{
rlllr
} (
x^3
&
+4x^2
&
+x
&
-6)
&
:
(x-1)
=\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6}
\\
\textcolor[rgb]{1,0,0}{{\underline{-(
x^3}}
&
\textcolor[rgb]{1,0,0}{\underline{-x^2)}}}
&
&
&
\\
0
&
+5x^2
&
+x
\\
&
\textcolor[rgb]{0,0,1}{\underline{-(5x^2}}
&
\textcolor[rgb]{0,0,1}{\underline{-5x)}}
\\
&
0
&
+6x
&
-6
\\
&&
\textcolor[rgb]{0,1,0}{\underline{-(6x}}
&
\textcolor[rgb]{0,1,0}{\underline{-6)}}
\\
&&
0
&
+0
\end{array}

so we have: $$x^3+4x^2+x-6=(x-1)(x^2+5x+6)$$

x^3+4x^2+x-6=(x-1)(x^2+5x+6)

We solve:

$$\\x^2+5x+6=0\\
x^2+5x+(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2-(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2+6=0\\
\left( x+\frac{5}{2}\right)^2=\frac{25}{4}-6\\
\left( x+\frac{5}{2}\right)^2=\frac{1}{4}\qquad| \quad \pm\sqrt{}\\
x+\frac{5}{2}=\pm\sqrt{\frac{1}{4}}\\
x_{2,3}=-\frac{5}{2}\pm\frac{1}{2}\\
x_2=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=\underline{-2}\\
x_3=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=\underline{-3}\\$$

\\x^2+5x+6=0\\
x^2+5x+(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2-(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2+6=0\\
\left( x+\frac{5}{2}\right)^2=\frac{25}{4}-6\\
\left( x+\frac{5}{2}\right)^2=\frac{1}{4}\qquad| \quad \pm\sqrt{}\\
x+\frac{5}{2}=\pm\sqrt{\frac{1}{4}}\\
x_{2,3}=-\frac{5}{2}\pm\frac{1}{2}\\
x_2=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=\underline{-2}\\
x_3=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=\underline{-3}\\

$$\\x_1=1 \qquad x_2=-2 \qquad x_3=-3\\
\boxed{x^3+4x^2+x-6=(x-1)(x+2)(x+3)}$$

\\x_1=1 \qquad x_2=-2 \qquad x_3=-3\\
\boxed{x^3+4x^2+x-6=(x-1)(x+2)(x+3)}

 
heureka May 15, 2014