Question

помогите срочняк $6cos2x-14cos^2x-7sin2x=0 \left[-3 \pi /2 - \pi /2]$

Answer 1

6cos2x-14cos²x-7sin2x=0
6(cos²x-sin²x)-14cos²x-7*2sinx*cosx=0
6cos²x-6sin²x-14cos²x-14sinx*cosx=0
-6sin²x-8cos²x-14sinx*cos=0  |: (-2cos²x)
3tg²x+4+7tgx=0
tgx=m
3m²+7m+4=0
m₁=-4/3
m₂=-1

tgx=-4/3, x₁=-actg(4/3)+ πn, n∈Z
tgx=-1, x₂= -arctg1+πn, n∈Z  x₂=-π/4+πn, n∈Z
x₁=-arctg4/3-π
x₂=-π/4-π