sin^2x+2sincosx-3cos^2x=0 | /cos^2x
tg^2x+2tgx-3=0
tgx=y
y^2+2y-3=0
y=1
y=-3
Найдем х :
1)tgx=1
x=pi/4+pik . k=z
2)tgx=-3
x=arctg(-3)+pik . k=z
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sinx/2+cosx/2+sinx/2cosx/2=1 |/cosx/2
tgx/2+1+sinx/2=1
tgx/2+sinx/2=0
(sinx/2+sinx/2*cosx/2)/cosx/2=0
одз:
1)sinx/2+sinx/2*cosx/2=0
sinx/2(1+cosx/2)=0
1.sinx/2=0
x/2=pik
x=2pik. k=z
2.cosx/2=-1
x/2=pi/2+pik
x=pi+2pik . k=z
2)cosx/2≠0
x≠pi/2+pik . k=z
Ответ:x=pik . k=z
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sinxcosx-sin^2x+sinx-cosx=0
sinx(cosx-sinx)-(cosx-sinx)=0
(cosx-sinx)(sinx-1)=0
1)cosx-sinx=0 |/sinx
ctgx=1
x=pi/4+pik . k=z
2)sinx-1=0
sinx=1
x=pi/2+2pik . k=z