Реши пожалуйста одно любое задание из 2-ой части

(2sin²x-7sinx+3)log14(-cosx)=0
2sin²x-7sinx+3=0                           или                      log14(-cosx)=0
t=sinx                                                                         ОДЗ: -cosx>0
2t²-7t+3=0                                                                            cosx<0
D=7²-4*2*3=49-24=25=5²                                                    x∈(π/2+2πn, 3π/2+2πn), n∈Z
t1=(7+5)/4=12/4=3                                                    log14(-cosx)=0
t2=(7-5)/4=2/4=1/2                                                    -cosx=14^0
sinx≠3, т.к. |sinx|≤1, а 3>1                                        -cosx=1
sinx=1/2  x=(-1)^k *π/6+πk, n∈Z                              cosx=-1
С учётом ОДЗ получаем:                                        x=π+2πn, n∈Z ( входит в ОДЗ)
x=5π/6+2πk, k∈Z
Ответ: 5π/6+2πk, π+2πn,  n,k∈Z