Question

3+sin2x=4sin^2x Решение

Answer 1

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