Решите уравнения 1) 2sinxcosx+5cos^2x=4 2) 3sin^2x-2sinxcosx=1

1) ы $sin2x+5((1+cos2x)/2)=4 (2sin2x+5+5cos2x)/2=4 2sin2x+5+5cos2x=8 2sin2x+5cos2x=3 2*(2tgx/1+tg^2x)+5(1-tg^2x/1+tg^2x)=3 4tgx+5-5tg^2x=3+3tg^2x 4tg^2x-2tgx-1=0 tgx=y 4y^2-2y-1=0$

2) $3-3cos2x/2-sin2x=1 3-3cos2x-2sin2x=2 3cos2x+2sin2x=1 3(1-tg^2x/1+tg^2x)+2(2tgx/1+tg^2x) 3-3tg^2x+4tgx=1+tg^2x 4tg^2x-4tgx-2=0 2tg^2x-2tgx-1=0 tgx=y 2y^2-2y-1=0$