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Peremand
@Peremand
August 2022
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cos(2x-3pi/2)=(√2)sinx
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boss9800
-sin2x=√2sinx
-2sinxcosx-√2sinx=0
-sinx(2cosx+√2)=0
-sinx=0 и 2cosx+√2=0
sinx=0 cosx=-√2/2
x=pi.n x=pi-pi/4+2pi.n
x=3pi/4+2pi.n
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Answers & Comments
-2sinxcosx-√2sinx=0
-sinx(2cosx+√2)=0
-sinx=0 и 2cosx+√2=0
sinx=0 cosx=-√2/2
x=pi.n x=pi-pi/4+2pi.n
x=3pi/4+2pi.n