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Question Number 145109 by mathdanisur last updated on 02/Jul/21
Solve the equation:  cos(6x)−cos(4x)=4y^2 +4y+3
$${Solve}\:{the}\:{equation}: \\ $$$${cos}\left(\mathrm{6}{x}\right)−{cos}\left(\mathrm{4}{x}\right)=\mathrm{4}{y}^{\mathrm{2}} +\mathrm{4}{y}+\mathrm{3} \\ $$
Answered by mitica last updated on 02/Jul/21
cos6x−cos4x≤1−(−1)=2⇒  4y^2 +4y+3≤2⇒(2y+1)^2 ≤0  ⇒y=((−1)/2)  cos4x=−1⇒cos2x=((1+cos4x)/2)=0  cos6x=1⇒4cos^3 2x−3cos2x=1⇒0=1⇒x∈φ
$${cos}\mathrm{6}{x}−{cos}\mathrm{4}{x}\leqslant\mathrm{1}−\left(−\mathrm{1}\right)=\mathrm{2}\Rightarrow \\ $$$$\mathrm{4}{y}^{\mathrm{2}} +\mathrm{4}{y}+\mathrm{3}\leqslant\mathrm{2}\Rightarrow\left(\mathrm{2}{y}+\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$\Rightarrow{y}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${cos}\mathrm{4}{x}=−\mathrm{1}\Rightarrow{cos}\mathrm{2}{x}=\frac{\mathrm{1}+{cos}\mathrm{4}{x}}{\mathrm{2}}=\mathrm{0} \\ $$$${cos}\mathrm{6}{x}=\mathrm{1}\Rightarrow\mathrm{4}{cos}^{\mathrm{3}} \mathrm{2}{x}−\mathrm{3}{cos}\mathrm{2}{x}=\mathrm{1}\Rightarrow\mathrm{0}=\mathrm{1}\Rightarrow{x}\in\phi \\ $$
Commented by mathdanisur last updated on 02/Jul/21
a lot cool Ser, thanks
$${a}\:{lot}\:{cool}\:{Ser},\:{thanks} \\ $$

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