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1 - cos(2x) + cos(6x) - cos(8x)

Guest May 7, 2015

Best Answer 

 #1
avatar+18715 
+15

1 - cos(2x) + cos(6x) - cos(8x) = 0       x =?

$$\small{\text{$
\begin{array}{rcl}
1-\cos{(2x)}+\cos{(6x)}-\cos{(8x)}&=&0\\
1-\cos{(2x)} &=&\cos{(8x)}-\cos{(6x)}\\
&&$Formula:\\
&&
$~\boxed{\mathbf{ \cos{2x}=1-2\sin^2{(x)} \quad or \quad 1-\cos{2x}=2\sin^2{(x)} }}\\
2\sin^2{(x)} &=& \cos{(8x)}-\cos{(6x)}\\
&&$Formula:\\
&&
$~\boxed{\mathbf{
\cos{(a)}-\cos{(b)}=2\sin{ \left( \dfrac{a+b}{2} \right)}\sin{ \left( \dfrac{b-a}{2} \right)}
}}\\
&&
\cos{(8x)}-\cos{(6x)}=2\sin{ \left( \dfrac{14x}{2} \right)}\sin{ \left( \dfrac{-2x}{2} \right)}\\
&&
\cos{(8x)}-\cos{(6x)}= -2\sin{ (7x) }\sin{(x)}\\
2\sin^2{(x)} &=& -2\sin{ (7x) }\sin{(x)}\\
\sin^2{(x)} &=& -\sin{ (7x) }\sin{(x)}\\
\sin^2{(x)} +\sin{ (7x) }\sin{(x)} &=& 0\\
\sin{(x)} \left[ \sin{(x)} + \sin{ (7x) } \right] &=& 0\\
&&$Formula:\\
&&
$~\boxed{\mathbf{
\sin{(a)}+\sin{(b)}=2\sin{ \left( \dfrac{a+b}{2} \right)}\cos{ \left( \dfrac{b-a}{2} \right)}
}}\\
&&
\sin{(x)}+\sin{(7x)}=2\sin{ \left( \dfrac{8x}{2} \right)}\cos{ \left( \dfrac{6x}{2} \right)}\\
&&
\sin{(x)}+\sin{(7x)}=2\sin{ (4x) }\cos{ (3x) }\\
\sin{(x)}\cdot 2 \cdot\sin{ (4x) }\cos{ (3x) } &=& 0\\
\end{array}
$}}$$

The solution set of the given equation is:

$$\\\boxed{\mathbf{
\sin{(x)}=0 \qquad \textcolor[rgb]{150,0,0}{x = k\cdot\pi \qquad k \in \mathrm{Z} }
}}\\
\boxed{\mathbf{
\sin{(4x)}=0 \qquad 4x = k\cdot \pi \qquad \textcolor[rgb]{150,0,0}{x = k\cdot \dfrac{\pi}{4} \qquad k \in \mathrm{Z} }
}}\\
\boxed{\mathbf{
\cos{(3x)}=0 \qquad 3x = \pm \dfrac{\pi}{2}+ k\cdot 2\cdot \pi \qquad \textcolor[rgb]{150,0,0}{x = \pm\dfrac{\pi}{6}
+ k\cdot \dfrac{2}{3}~\pi \qquad k \in \mathrm{Z} }
}}$$

heureka  May 8, 2015
Sort: 

8+0 Answers

 #1
avatar+18715 
+15
Best Answer

1 - cos(2x) + cos(6x) - cos(8x) = 0       x =?

$$\small{\text{$
\begin{array}{rcl}
1-\cos{(2x)}+\cos{(6x)}-\cos{(8x)}&=&0\\
1-\cos{(2x)} &=&\cos{(8x)}-\cos{(6x)}\\
&&$Formula:\\
&&
$~\boxed{\mathbf{ \cos{2x}=1-2\sin^2{(x)} \quad or \quad 1-\cos{2x}=2\sin^2{(x)} }}\\
2\sin^2{(x)} &=& \cos{(8x)}-\cos{(6x)}\\
&&$Formula:\\
&&
$~\boxed{\mathbf{
\cos{(a)}-\cos{(b)}=2\sin{ \left( \dfrac{a+b}{2} \right)}\sin{ \left( \dfrac{b-a}{2} \right)}
}}\\
&&
\cos{(8x)}-\cos{(6x)}=2\sin{ \left( \dfrac{14x}{2} \right)}\sin{ \left( \dfrac{-2x}{2} \right)}\\
&&
\cos{(8x)}-\cos{(6x)}= -2\sin{ (7x) }\sin{(x)}\\
2\sin^2{(x)} &=& -2\sin{ (7x) }\sin{(x)}\\
\sin^2{(x)} &=& -\sin{ (7x) }\sin{(x)}\\
\sin^2{(x)} +\sin{ (7x) }\sin{(x)} &=& 0\\
\sin{(x)} \left[ \sin{(x)} + \sin{ (7x) } \right] &=& 0\\
&&$Formula:\\
&&
$~\boxed{\mathbf{
\sin{(a)}+\sin{(b)}=2\sin{ \left( \dfrac{a+b}{2} \right)}\cos{ \left( \dfrac{b-a}{2} \right)}
}}\\
&&
\sin{(x)}+\sin{(7x)}=2\sin{ \left( \dfrac{8x}{2} \right)}\cos{ \left( \dfrac{6x}{2} \right)}\\
&&
\sin{(x)}+\sin{(7x)}=2\sin{ (4x) }\cos{ (3x) }\\
\sin{(x)}\cdot 2 \cdot\sin{ (4x) }\cos{ (3x) } &=& 0\\
\end{array}
$}}$$

The solution set of the given equation is:

$$\\\boxed{\mathbf{
\sin{(x)}=0 \qquad \textcolor[rgb]{150,0,0}{x = k\cdot\pi \qquad k \in \mathrm{Z} }
}}\\
\boxed{\mathbf{
\sin{(4x)}=0 \qquad 4x = k\cdot \pi \qquad \textcolor[rgb]{150,0,0}{x = k\cdot \dfrac{\pi}{4} \qquad k \in \mathrm{Z} }
}}\\
\boxed{\mathbf{
\cos{(3x)}=0 \qquad 3x = \pm \dfrac{\pi}{2}+ k\cdot 2\cdot \pi \qquad \textcolor[rgb]{150,0,0}{x = \pm\dfrac{\pi}{6}
+ k\cdot \dfrac{2}{3}~\pi \qquad k \in \mathrm{Z} }
}}$$

heureka  May 8, 2015
 #2
avatar+91049 
+5

Like WOW Heureka,    

MathsGod1 is learning LaTex.  He is only 12 but he is learning very quickly.

I should send him to you  :))

I'm quite happy to keep teaching him but you can if you would like to.

You are Lord of LaTex in Camelot. :))  

Melody  May 9, 2015
 #3
avatar+78750 
0

Very nice, heureka....!!!!

 

  

CPhill  May 9, 2015
 #4
avatar+4663 
+5

aha Melody,  I've noticed Heureka's LaTeX ever sinced i layed my eyes on this website and LaTeX.

 

The LaTeX is amazing but i think i should take it slow before i get to your level...Really slow :D

MathsGod1  May 9, 2015
 #5
avatar+78750 
0

Your LaTex abilities are improving all the time, MG1....!!!!

 

Maybe....one day......you can teach me....LOL  !!!!!

 

 

  

CPhill  May 9, 2015
 #6
avatar+4663 
+5

Hahah!

 

The first time i teach a grown up.

Lol.

 

Ok, we'll trade in exchange for LaTeX give me 99% of your points!

 

 

Just joking...We'll need someone like you as Moderator I'm not sure I'll do great.

 

 

LaTeX looks hard but once leart it actually isn't.

Just a whole bunch of numbers, letters and symbols!

MathsGod1  May 9, 2015
 #7
avatar+78750 
0

.......Just a whole bunch of numbers, letters and symbols!.....

 

Hey......that kinda' sounds like.......MATH....!!!!

 

 

  

CPhill  May 9, 2015
 #8
avatar+4663 
0

That is pure logic.

MathsGod1  May 10, 2015

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