Question Number 203352 by Mingma last updated on 17/Jan/24
Answered by witcher3 last updated on 17/Jan/24
$$\mathrm{sin}\left(\mathrm{2x}\right)=\mathrm{0}\Rightarrow\mathrm{x}=\frac{\mathrm{k}\pi}{\mathrm{2}},\mathrm{k}\in\mathbb{Z} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{4}=\mathrm{x}^{\mathrm{2}} −\left(\mathrm{1}+\mathrm{4}\right)\mathrm{x}+\left(\mathrm{1}.\mathrm{4}\right)=\mathrm{0}\Rightarrow\mathrm{x}\in\left\{\mathrm{1},\mathrm{4}\:\right\} \\ $$$$\mathrm{X}=\mathbb{R}−\left(\left\{\mathrm{1},\mathrm{4}\right\}\cup\left\{\frac{\mathrm{k}\pi}{\mathrm{2}},\mathrm{k}\in\mathbb{Z}\right\}\right) \\ $$
Commented by Mingma last updated on 17/Jan/24
Perfect ��