Question

cele două ecuații!!!​

Answer 1

Răspuns:

Substitutie\

π/3-x=π/2-(x+π/6)

cos[π/2-(x+π/6)]=sin (x+π/6)

sin(x+π/6)=y   y∈[-1,1]

3y²-y=1

--

3y²-y-1=0

y1=(1-√13)/2<-1  nu  convine  problemei

y2=(1+√13)/2>1 u convine  probklemei

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π/4-x =tg(π/2-(π/4+x)

3tg(π/4+x)+2tg[π/2-(π/4+x)]=5

3tg(π/4+x)+2ctg(π/4+x)]=5

3tg(π/4+x)+2/tg(π/4+x)=5

tg (π/4+x)=y

3y+2/y=5

3y²-5y+2=0

y1=1=> tgx=1=> x=π/4

Solutia  generala  x=kπ±π/4

y2=2/3 tgx=2/3  x  =arctg 2/3

Solutuia  generala x2=arctg 2/3±kπ

In finaLx1Ux2

Explicație pas cu pas: