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Question Number 73538 by Rio Michael last updated on 13/Nov/19

find the general solution of sin4x + cos2x = 0

$${find}\:{the}\:{general}\:{solution}\:{of}\: \\ $$$$\:{sin}\mathrm{4}{x}\:+\:{cos}\mathrm{2}{x}\:=\:\mathrm{0} \\ $$

Answered by ajfour last updated on 13/Nov/19

cos 2x(2sin 2x+1)=0 ⇒ 2x=(2n+1)(π/2) or 2x=(4n−1)(π/2)±(π/3) .

$$\mathrm{cos}\:\mathrm{2}{x}\left(\mathrm{2sin}\:\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{2}{x}=\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\:\:\:{or} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{x}=\left(\mathrm{4}{n}−\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\pm\frac{\pi}{\mathrm{3}}\:\:. \\ $$

Commented by Rio Michael last updated on 13/Nov/19

thanks sir

$${thanks}\:{sir} \\ $$

Answered by malwaan last updated on 14/Nov/19

2sin2x cos2x + cos2x =0 cos2x(2sin2x+1)=0 cos2x=0⇒2x=(𝛑/2) + 2n𝛑 ⇒x=(𝛑/4) + n𝛑 or 2sin2x+1=0 ⇒2sin2x=−1⇒sin2x=−(1/2) ⇒2x=((𝛑/6)+𝛑)+2n𝛑=((7𝛑)/6)+2n𝛑 ⇒x=((7𝛑)/(12)) + n𝛑 or 2x=(2𝛑−(𝛑/6))+2n𝛑 =((11𝛑)/6) +2n𝛑 ⇒x=((11𝛑)/(12)) + n𝛑 ⇒x={((𝛑/4)∨((7𝛑)/(12))∨((11𝛑)/(12)))+n𝛑}

$$\mathrm{2}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}\:\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\:+\:\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\:=\mathrm{0} \\ $$$$\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\left(\mathrm{2}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}=\mathrm{0}\Rightarrow\mathrm{2}\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\:+\:\mathrm{2}\boldsymbol{{n}\pi} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{4}}\:+\:\boldsymbol{{n}\pi} \\ $$$$\boldsymbol{{or}}\:\mathrm{2}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}=−\mathrm{1}\Rightarrow\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\boldsymbol{{x}}=\left(\frac{\boldsymbol{\pi}}{\mathrm{6}}+\boldsymbol{\pi}\right)+\mathrm{2}\boldsymbol{{n}\pi}=\frac{\mathrm{7}\boldsymbol{\pi}}{\mathrm{6}}+\mathrm{2}\boldsymbol{{n}\pi} \\ $$$$\Rightarrow\boldsymbol{{x}}=\frac{\mathrm{7}\boldsymbol{\pi}}{\mathrm{12}}\:+\:\boldsymbol{{n}\pi} \\ $$$$\boldsymbol{{or}}\:\mathrm{2}\boldsymbol{{x}}=\left(\mathrm{2}\boldsymbol{\pi}−\frac{\boldsymbol{\pi}}{\mathrm{6}}\right)+\mathrm{2}\boldsymbol{{n}\pi} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{11}\boldsymbol{\pi}}{\mathrm{6}}\:+\mathrm{2}\boldsymbol{{n}\pi} \\ $$$$\Rightarrow\boldsymbol{{x}}=\frac{\mathrm{11}\boldsymbol{\pi}}{\mathrm{12}}\:+\:\boldsymbol{{n}\pi} \\ $$$$\Rightarrow\boldsymbol{{x}}=\left\{\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\vee\frac{\mathrm{7}\boldsymbol{\pi}}{\mathrm{12}}\vee\frac{\mathrm{11}\boldsymbol{\pi}}{\mathrm{12}}\right)+\boldsymbol{{n}\pi}\right\} \\ $$

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