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Question Number 92461 by jagoll last updated on 07/May/20

cos (((3x)/4)−π)=sin ((π/4)−2x)  x ∈ [0, π ]

$$\mathrm{cos}\:\left(\frac{\mathrm{3x}}{\mathrm{4}}−\pi\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{2x}\right) \\ $$$$\mathrm{x}\:\in\:\left[\mathrm{0},\:\pi\:\right]\: \\ $$

Commented by john santu last updated on 07/May/20

−cos(((3x)/4)) =sin ((π/4)−2x)   cos (((3x)/4)) = sin (2x−(π/4))  cos (((3x)/4)) = cos ((π/2)+(π/4)−2x)   { ((((11x)/4) = ((3π)/4)+ 2kπ)),((−((5x)/4) = −((3π)/4)+2kπ)) :}   { ((x=((3π)/(11))+((8kπ)/(11)))),((x=((3π)/5)−((8kπ)/5))) :}

$$−\mathrm{cos}\left(\frac{\mathrm{3x}}{\mathrm{4}}\right)\:=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{2x}\right)\: \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{3x}}{\mathrm{4}}\right)\:=\:\mathrm{sin}\:\left(\mathrm{2x}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{3x}}{\mathrm{4}}\right)\:=\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}−\mathrm{2x}\right) \\ $$$$\begin{cases}{\frac{\mathrm{11x}}{\mathrm{4}}\:=\:\frac{\mathrm{3}\pi}{\mathrm{4}}+\:\mathrm{2k}\pi}\\{−\frac{\mathrm{5x}}{\mathrm{4}}\:=\:−\frac{\mathrm{3}\pi}{\mathrm{4}}+\mathrm{2k}\pi}\end{cases} \\ $$$$\begin{cases}{\mathrm{x}=\frac{\mathrm{3}\pi}{\mathrm{11}}+\frac{\mathrm{8k}\pi}{\mathrm{11}}}\\{\mathrm{x}=\frac{\mathrm{3}\pi}{\mathrm{5}}−\frac{\mathrm{8k}\pi}{\mathrm{5}}}\end{cases} \\ $$

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