Question Number 92461 by jagoll last updated on 07/May/20
![cos (((3x)/4)−π)=sin ((π/4)−2x) x ∈ [0, π ]](https://www.tinkutara.com/question/Q92461.png)
$$\mathrm{cos}\:\left(\frac{\mathrm{3x}}{\mathrm{4}}−\pi\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{2x}\right) \\ $$$$\mathrm{x}\:\in\:\left[\mathrm{0},\:\pi\:\right]\: \\ $$
Commented by john santu last updated on 07/May/20
$$−\mathrm{cos}\left(\frac{\mathrm{3x}}{\mathrm{4}}\right)\:=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{2x}\right)\: \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{3x}}{\mathrm{4}}\right)\:=\:\mathrm{sin}\:\left(\mathrm{2x}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{3x}}{\mathrm{4}}\right)\:=\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}−\mathrm{2x}\right) \\ $$$$\begin{cases}{\frac{\mathrm{11x}}{\mathrm{4}}\:=\:\frac{\mathrm{3}\pi}{\mathrm{4}}+\:\mathrm{2k}\pi}\\{−\frac{\mathrm{5x}}{\mathrm{4}}\:=\:−\frac{\mathrm{3}\pi}{\mathrm{4}}+\mathrm{2k}\pi}\end{cases} \\ $$$$\begin{cases}{\mathrm{x}=\frac{\mathrm{3}\pi}{\mathrm{11}}+\frac{\mathrm{8k}\pi}{\mathrm{11}}}\\{\mathrm{x}=\frac{\mathrm{3}\pi}{\mathrm{5}}−\frac{\mathrm{8k}\pi}{\mathrm{5}}}\end{cases} \\ $$