Тригонометрично уравнение

[Гост]

[tex]cosx.cos2x.cos4x=\frac{1}{8}[/tex]

[Гост]

[tex]cosx.cos2x.cos4x=\frac{1}{8}[/tex]
[tex]\Rightarrow 8cosx.cos2x.cos4x=1[/tex]
[tex]\Rightarrow 8cosx.cos2x.cos4x.sinx=sinx ;(sinx \ne 0 \Rightarrow x \ne k\pi)[/tex]
[tex]\Rightarrow 4sin2x.cos2x.cos4x=sinx[/tex]
[tex]\Rightarrow 2sin4x.cos4x=sinx[/tex]
[tex]\Rightarrow sin8x=sinx[/tex]
[tex]\Rightarrow sin8x-sinx=0[/tex]
[tex]\Rightarrow 2sin {\frac{8x-x}{2}}.cos {\frac{8x+x}{2}}=0[/tex]

[tex]\Rightarrow \frac{7x}{2}=k_1\pi \Rightarrow x=\frac{2k_1\pi}{7} (k_1 \ne 7l_1)[/tex]
[tex]\Rightarrow \frac{9x}{2}=\frac{\pi}{2}+k_2\pi \Rightarrow x=\frac{\pi}{9}+\frac{2k_2\pi}{9} (k_2 \ne 9l_2+4)[/tex]

[Гост]

Благодря ти, Гост. Само искам да попитам как намери [tex]k_{1}\ne 7l_{1}[/tex] и [tex]k_{2}\ne 9l_{2}+4[/tex]
Аз предполагам ,че защото [tex]sinx\ne 0 , x\ne k\pi[/tex]
[tex]x=\frac{2k_{1}\pi}{7}\ne l_{1}\pi[/tex]
[tex]2k_{1}\cancel{\pi}\ne 7l_{1}\cancel{\pi}[/tex]
[tex]2k_{1}\ne 7l_{1}[/tex]
[tex]k_{1}\ne \frac{7l_{1}}{2}[/tex], но понеже [tex]k_{1} \in Z[/tex], трябва да е цяло и затова ли [tex]k_{1}\ne 7l_{1}[/tex]

[tex]x=\frac{\pi}{9}+\frac{2k_{2}\pi}{9}\ne l_{2}\pi[/tex]
[tex]\frac{\cancel{\pi}}{9}(2k_{2}+1)\ne l_{2}\cancel{\pi}[/tex]
[tex]2k_{2}+1\ne l_{2}[/tex]
[tex]k_{2}\ne \frac{9l_{2}-1}{2}[/tex]
HELP

[Гост]

..........
[tex]x=\frac{\pi}{9}+\frac{2k_{2}\pi}{9}\ne l_{2}\pi[/tex]
[tex]\frac{\cancel{\pi}}{9}(2k_{2}+1)\ne l_{2}\cancel{\pi}[/tex]
[tex]2k_{2}+1\ne l_{2}[/tex]
[tex]k_{2}\ne \frac{9l_{2}-1}{2}[/tex]
HELP

[tex]k_{2}\ne \frac{9l_{2}-1}{2} \Rightarrow l_2=2l_3+1 \Rightarrow k_{2}\ne \frac{9(2l_{3}+1)-1}{2} \Rightarrow k_2 \ne 9l_3+4[/tex]