Question Number 146725 by mathdanisur last updated on 15/Jul/21
$${cos}\left({x}\right)\:\centerdot\:{cos}\left(\mathrm{3}{x}\right)\:=\:{cos}\left(\mathrm{5}{x}\right)\:\centerdot\:{cos}\left(\mathrm{7}{x}\right) \\ $$$$\Rightarrow\:{x}\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 15/Jul/21
![cosx.cos3x = cos5x.cos7x (1/2)[cos(x−3x)+cos(x+3x)] = (1/2)[cos(5x−7x)+cos(5x+7x)] cos4x = cos12x 12x = ±4x+2kπ { ((8x = 2kπ)),((16x = 2kπ)) :} { ((x = ((kπ)/4))),((x = ((kπ)/8))) :}](https://www.tinkutara.com/question/Q146727.png)
$$\mathrm{cos}{x}.\mathrm{cos3}{x}\:=\:\mathrm{cos5}{x}.\mathrm{cos7}{x} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\left({x}−\mathrm{3}{x}\right)+\mathrm{cos}\left({x}+\mathrm{3}{x}\right)\right]\:= \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\left(\mathrm{5}{x}−\mathrm{7}{x}\right)+\mathrm{cos}\left(\mathrm{5}{x}+\mathrm{7}{x}\right)\right] \\ $$$$ \\ $$$$\mathrm{cos4}{x}\:=\:\mathrm{cos12}{x} \\ $$$$\mathrm{12}{x}\:=\:\pm\mathrm{4}{x}+\mathrm{2}{k}\pi \\ $$$$\begin{cases}{\mathrm{8}{x}\:=\:\mathrm{2}{k}\pi}\\{\mathrm{16}{x}\:=\:\mathrm{2}{k}\pi}\end{cases} \\ $$$$\begin{cases}{{x}\:=\:\frac{{k}\pi}{\mathrm{4}}}\\{{x}\:=\:\frac{{k}\pi}{\mathrm{8}}}\end{cases} \\ $$
Commented by mathdanisur last updated on 15/Jul/21
$${thank}\:{you}\:{Ser} \\ $$
Answered by bramlexs22 last updated on 15/Jul/21
$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\mathrm{4x}+\mathrm{cos}\:\mathrm{2x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\mathrm{12x}+\mathrm{cos}\:\mathrm{2x}\right) \\ $$$$\Leftrightarrow\mathrm{cos}\:\mathrm{4x}+\mathrm{cos}\:\mathrm{2x}=\mathrm{cos}\:\mathrm{12x}+\mathrm{cos}\:\mathrm{2x} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{12x}−\mathrm{cos}\:\mathrm{4x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2cos}\:\mathrm{8x}\:\mathrm{cos}\:\mathrm{4x}=\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{cos}\:\mathrm{8x}=\mathrm{0}\Rightarrow\mathrm{8x}=\mathrm{2n}\pi\pm\frac{\pi}{\mathrm{2}}}\\{\mathrm{cos}\:\mathrm{4x}=\mathrm{0}\Rightarrow\mathrm{4x}=\mathrm{2n}\pi\pm\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$
Commented by mathdanisur last updated on 15/Jul/21
$${thank}\:{you}\:{Ser} \\ $$