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Question Number 143105 by mathdanisur last updated on 10/Jun/21

((cos(3x))/(sin(2x))) = 0

$$\frac{{cos}\left(\mathrm{3}{x}\right)}{{sin}\left(\mathrm{2}{x}\right)}\:=\:\mathrm{0} \\ $$

Commented by Dwaipayan Shikari last updated on 10/Jun/21

cos(3x)=0     3x=(2k+1)(π/2)⇒x=(2k+1)(π/6)    k∈Z  sin(2x)≠0  2x≠gπ⇒x≠((gπ)/2)  (2k+1)(π/6)≠((gπ)/2)⇒2k+1≠3g⇒((2k+1)/3)≠g

$${cos}\left(\mathrm{3}{x}\right)=\mathrm{0}\:\:\: \\ $$$$\mathrm{3}{x}=\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\Rightarrow{x}=\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{6}}\:\:\:\:{k}\in\mathbb{Z} \\ $$$${sin}\left(\mathrm{2}{x}\right)\neq\mathrm{0} \\ $$$$\mathrm{2}{x}\neq{g}\pi\Rightarrow{x}\neq\frac{{g}\pi}{\mathrm{2}} \\ $$$$\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{6}}\neq\frac{{g}\pi}{\mathrm{2}}\Rightarrow\mathrm{2}{k}+\mathrm{1}\neq\mathrm{3}{g}\Rightarrow\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{3}}\neq{g} \\ $$

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