Question

sa se rezolve in [0,2pi) ecuatia sinx+cosx=-1

Answer 1

Răspuns:

Explicație pas cu pas:

$sinx+cosx+1=0,~~sin(2*\frac{x}{2})+cos(2*\frac{x}{2})+1=0,~~2sin\frac{x}{2}cos\frac{x}{2}+cos^{2}\frac{x}{2}-sin^{2}\frac{x}{2}+sin^{2}\frac{x}{2}+cos^{2}\frac{x}{2}=0,~~ 2sin\frac{x}{2}cos\frac{x}{2}+2cos^{2}\frac{x}{2} =0,~~2cos\frac{x}{2}(sin\frac{x}{2}+cos\frac{x}{2})=0\\ Deci,~cos\frac{x}{2}=0~~sau~~ sin\frac{x}{2}+cos\frac{x}{2}=0\\1)~cos\frac{x}{2}=0,~=>~\frac{x}{2}=\frac{\pi }{2}+\pi k~|*2,~=>~x=\pi +2\pi k,\\pentru~k=0,~x=\pi ~apartine~[0,2\pi ]$

$2)~sin\frac{x}{2}+cos\frac{x}{2}=0~|:cos\frac{x}{2},~~=>~tg\frac{x}{2}+1=0,~=>~ tg\frac{x}{2}=-1,~=>~\frac{x}{2}=\frac{3\pi}{4} +\pi k~|*2,~=>~x=\frac{3\pi }{2}+2\pi k.\\Pentru~k=0,~x=\frac{3\pi }{2},~apartine~[0,2\pi ]\\Raspuns:~~\pi ,~~\frac{3\pi }{2}.$