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justdanceandsing
@justdanceandsing
July 2022
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1) sin 4x=sin 2x
2) sin 2x=cos^2 x
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uekmyfhfp
1) 2 sin 2x cos 2x = sin 2x;
sin2x(2 cos2x -1)=0;
sin2x=0; ⇒2x= pik; x= pk/2; k-Z,
2 cos2x - 1 =0;
cos 2x= 1/2;
2x= + - pi/3 + 2pin;
x = + - pi/6 + pin; n-Z.
2) 2 sinxcosx=cos^2x
2 sinxcosx - cos^2x =0;
cosx(2 sinx - cosx)=0;
cosx=0;⇒x=pi/2 + pik; k-Z.
2 sinx - cos x=0;
2 sinx= cos x; / 2 cos x≠0;
tgx=1/2;
x= arctg1/2 + pin; n-Z
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Answers & Comments
sin2x(2 cos2x -1)=0;
sin2x=0; ⇒2x= pik; x= pk/2; k-Z,
2 cos2x - 1 =0;
cos 2x= 1/2;
2x= + - pi/3 + 2pin;
x = + - pi/6 + pin; n-Z.
2) 2 sinxcosx=cos^2x
2 sinxcosx - cos^2x =0;
cosx(2 sinx - cosx)=0;
cosx=0;⇒x=pi/2 + pik; k-Z.
2 sinx - cos x=0;
2 sinx= cos x; / 2 cos x≠0;
tgx=1/2;
x= arctg1/2 + pin; n-Z