Question

sin3x + sinx = 0

3sin²x + 3sinx cosx + 2cos²x = 1

√3 - 2sin x < 0

решите уравнения неравенства​

Answer 1

3sin²x+3sinxcosx+2cos²x-sin²x-cos²x=0

2sin²x+3sinxcosx+cos²x=0 /cos²x≠0

2tg²x+3tgx+1=0

D=9-8=1

a1=(-3-1)/4=-1 tgx=-1⇒x=-π/4+πn

a2=(-3+1)/4=-1/2 tgx=-1/2⇒x=-arctg1/2+πn