2sinx-3cosx=0 и cosx+sinx=cos2x заранее спасибо

2sinx-3cosx=0tg (x)=3/2x=artg (3/2) x=56,03 градусаcos (x)+sin (x)=2cos (2x) cos (x)+sin (x)=cos^2 (x) -sin^2 (x) cos (x)+sin (x)=(cos (x) -sin (x) (cos (x)+sin (x) (cos (x) -sin (x) -1) (cos (x)+sin (x)=0cos (x)+sin (x)=0tg (x)=-1 x=-п/4+2 пкcos (x) -sin (x) -1=0cos^2 (x)=1+2sin (x)+sin^2 (x) 1-sin^2 (x)=1+2sin (x)+sin^2 (x) 2sin (x)+2sin^2 (x)=02sin (x) (1+sin (x)=0sin (x)=0x=0+2 пк 1+sin (x)=0sin (x)=-1x=-п/2+2 пк