Question

sin^3x+sinx*sin2x-3 sin x*cos^x-6 cos^3x=0

Answer 1

sin2x=2sinxcosx  ⇒

sin³x+2sin²xcosx-3sinxcos²x-6cos³x=0

Делим ур-ие на cos³x≠0

tg³x+2tg²x-3tgx-6=0

tg²x(tgx+2)-3(tgx+2)=0

(tgx+2)(tg²x-3)=0,  (tgx+2)(tgx-√3)(tgx+√3)=0

a) tgx=-2, x= -arctg2+πn, n∈Z

b)  tgx=√3, x=π/3+πk, k∈Z

c)  tgx=-√3,  x= -π/3+πm, m∈Z