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