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vischneowa
@vischneowa
July 2022
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[tex]4sin^{2}2x+3cos4x-1=0 [/tex]
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Хильмилли
Cos4x=cos^2(2x)-sin^2(2x)=1-2sin^2(2x)⇒
4sin^2(2x)+3*(1-2sin^2(2x))-1=0⇒4sin^2(2x)+3-6sin^2(2x)-1=0⇒
-2sin^2(2x)+2=0⇒2sin^2(2x)=2⇒sin^2(2x)=1⇒
1) sin2x=1⇒2x=π/2+2πn⇒x=π/4+πn
2) sin2x=-1⇒2x=-π/2+2πn⇒x=-π/4+πn
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Answers & Comments
4sin^2(2x)+3*(1-2sin^2(2x))-1=0⇒4sin^2(2x)+3-6sin^2(2x)-1=0⇒
-2sin^2(2x)+2=0⇒2sin^2(2x)=2⇒sin^2(2x)=1⇒
1) sin2x=1⇒2x=π/2+2πn⇒x=π/4+πn
2) sin2x=-1⇒2x=-π/2+2πn⇒x=-π/4+πn