Question

2sinx+sin2x=cosx+1, x∈[2p/3;p) найдите корни уравнения

Answer 1

2sinx(1+cosx)-(cosx+1)=0
(cosx+1)(2sinx-1)=0
cosx=-1⇒x=π+2πn
sinx=1/2⇒x=(-1)^n*π/6+πn
x=5π/6

Answer 2

2sinx+2sinxcosx=cosx+1
2sinx(1+cosx)=cosx+1
2sinx=1
sinx=1/2
x=п/6