Răspuns:
x=-π/6 +2πk, k∈Z
Explicație pas cu pas:
[tex]2sin(x+\frac{\pi }{6})+\sqrt{3}sinx+cosx=0~~2(sinx*cos\frac{\pi }{6} +cosx*sin\frac{\pi }{6})+\sqrt{3}sinx+cosx=0~~2sinx*\frac{\sqrt{3} }{2}+2cosx*\frac{1}{2} \sqrt{3}sinx+cosx=0~~\sqrt{3}sinx+cosx+\sqrt{3}sinx+cosx=0~~2(\sqrt{3}sinx+cosx)=0 ~~\sqrt{3}sinx+cosx=0~|:cosx~~~\sqrt{3}tgx+1=0~~tgx=-\frac{1}{\sqrt{3} } \\[/tex]
Deoarece cosx>0 ⇒x=-π/6 +2πk, k∈Z.
