Question Number 148706 by mathdanisur last updated on 30/Jul/21
Answered by mindispower last updated on 30/Jul/21
$${sin}^{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}} \\ $$$$\left({E}\right)\Leftrightarrow\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{6}{x}\right)+{cos}\left(\mathrm{8}{x}\right)+{cos}\left(\mathrm{12}{x}\right)+{cos}\left(\mathrm{14}{x}\right)\right)=\mathrm{2} \\ $$$$\Leftrightarrow{cos}\left(\mathrm{6}{x}\right)+{cos}\left(\mathrm{14}{x}\right)+{cos}\left(\mathrm{8}{x}\right)+{cos}\left(\mathrm{12}{x}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2}{cos}\left(\mathrm{10}{x}\right){cos}\left(\mathrm{4}{x}\right)+\mathrm{2}{cos}\left(\mathrm{10}{x}\right){cos}\left(\mathrm{2}{x}\right)=\mathrm{0} \\ $$$$\Leftrightarrow{cos}\left(\mathrm{10}{x}\right)\left({cos}\left(\mathrm{4}{x}\right)+{cos}\left(\mathrm{2}{x}\right)\right)=\mathrm{0} \\ $$$$\Leftrightarrow{cos}\left(\mathrm{10}{x}\right){cos}\left(\mathrm{3}{x}\right){cos}\left({x}\right)=\mathrm{0} \\ $$$${x}=\frac{\pi}{\mathrm{2}}+{k}\pi \\ $$$${x}=\frac{\pi}{\mathrm{6}}+\frac{{k}\pi}{\mathrm{3}},{x}=\frac{\pi}{\mathrm{20}}+\frac{{k}\pi}{\mathrm{10}},{k}\in\mathbb{Z} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 30/Jul/21
$${Thank}\:{you}\:{Ser} \\ $$
Commented by mindispower last updated on 30/Jul/21
$${withe}\:{pleasur} \\ $$