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Find x if cosx = cos[180-cosinverse((0.25*cosx)/6.76)]

Guest May 15, 2014

Best Answer 

 #4
avatar+19597 
+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
 #1
avatar+92751 
+5

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{x}}\right)} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{180}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{0.25}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{x}}\right)}}{{\mathtt{6.76}}}}\right)}\right)} \Rightarrow \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{180}}}}\right)} = {\mathtt{\,-\,}}{\frac{{\mathtt{25}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{180}}}}\right)}}{{\mathtt{676}}}}$$

Umm that doesn't look very helpful.

Okay lets try something else

http://www.wolframalpha.com/input/?i=cos%28x%29%3Dcos%28180-acos%280.25%2F6.76*cos%28x%29%29%29

I'll think about it some more and i am sure others will as well.

Melody  May 15, 2014
 #2
avatar+19597 
+5

cosx = cos[180-cosinverse((0.25*cosx)/6.76)]        x in rad!!!

 

The Solution for  $$\cos{(x)} = \cos{(180\textcolor[rgb]{1,0,0}{rad}-cos^{-1}{(\;(0.25*\cos{(x)})/6.76})\;)}$$      x in rad is:

\cos{(x)} = \cos{(180\textcolor[rgb]{1,0,0}{rad}-cos^{-1}{(\;(0.25*\cos{(x)})/6.76})\;)}  

$$x=\pm cos^{-1}
\left(
\;\frac
{
\sin{ ( 180\;\textcolor[rgb]{1,0,0}{rad} ) }
}
{
\sqrt{
\left( \frac{0.25}{6.76} - \cos{
( 180\;\textcolor[rgb]{1,0,0}{rad} ) }
\right) ^2
+ ( \sin{ (180\;\textcolor[rgb]{1,0,0}{rad}) } )^2
}
}
\;\right)$$

x=\pm cos^{-1}
\left(
    \;\frac
{
  \sin{ ( 180\;\textcolor[rgb]{1,0,0}{rad} ) }
}
{
\sqrt{
     \left( \frac{0.25}{6.76} - \cos{
( 180\;\textcolor[rgb]{1,0,0}{rad} ) }
     \right) ^2
+ ( \sin{ (180\;\textcolor[rgb]{1,0,0}{rad}) }  )^2
}
}
\;\right) 

 

$$\\x=\pm (2.47103630997\pm 2\pi*k)\\\\
Example:\\
x=-3.81215\\
x=-2.47104\\
x=2.47104\\
x=3.81245\\
\dots$$

\\x=\pm (2.47103630997\pm 2\pi*k)\\\\
Example:\\
x=-3.81215\\
x=-2.47104\\
x=2.47104\\
x=3.81245\\
\dots

heureka  May 15, 2014
 #3
avatar+26741 
+5

Hmm!  Unusual to have 180 radians.  What if the 180 is 180°, or $$\pi$$ radians? In this case we get a different set of results:

coscos

The values of x are now given by $$x=(n+\frac{1}{2})\pi$$ where n is an integer.

Alan  May 15, 2014
 #4
avatar+19597 
+5
Best Answer

...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

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