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

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

Guest May 7, 2015

#1
+18827
+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:

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

heureka  May 8, 2015
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#1
+18827
+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:

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

heureka  May 8, 2015
#2
+91436
+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
+80875
0

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

CPhill  May 9, 2015
#4
+4664
+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
+80875
0

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

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

CPhill  May 9, 2015
#6
+4664
+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
+80875
0

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

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

CPhill  May 9, 2015
#8
+4664
0

That is pure logic.

MathsGod1  May 10, 2015

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