Question

3 + sin 2x = 4sin^2x

Answer 1

3+sin2x=4sin^2xsin2x=2sinx*cosx3+2sinx*cosx=4sin^2x  Делим на sin^2x3/sin^2x + 2ctgx =41/sin^2x=1+ctg^2x3(1+ctg^2x)+2ctgx=43ctg^2x+2ctgx-1=0D=4+12=16ctgx=1/3  -> x=arcctg 1/3 + pi*kctgx=-1 ->x=3pi/4+pi*l