+26396 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} $}}
The solution set of the given equation is:
\\\boxed{\mathbf{ \sin{(x)}=0 \qquad \textcolor[rgb]{150,0,0}{x = k\cdot\pi \qquad k \in \mathrm{Z} } }}\\ \boxed{\mathbf{ \sin{(4x)}=0 \qquad 4x = k\cdot \pi \qquad \textcolor[rgb]{150,0,0}{x = k\cdot \dfrac{\pi}{4} \qquad k \in \mathrm{Z} } }}\\ \boxed{\mathbf{ \cos{(3x)}=0 \qquad 3x = \pm \dfrac{\pi}{2}+ k\cdot 2\cdot \pi \qquad \textcolor[rgb]{150,0,0}{x = \pm\dfrac{\pi}{6} + k\cdot \dfrac{2}{3}~\pi \qquad k \in \mathrm{Z} } }}
+26396 1 - cos(2x) + cos(6x) - cos(8x) = 0 x =?
\small{\text{$ \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} } }}
+118696 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. :))
+4711 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
+130458 Your LaTex abilities are improving all the time, MG1....!!!!
Maybe....one day......you can teach me....LOL !!!!!
+4711 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!
