Question

Determinati solutiile ecuatiei: radical din 3*sinx-cosx=1 , x apartine [0,2pi]

Answer 1

Răspuns:

pi/3

Explicație pas cu pas:

$\sqrt{3} sinx-cosx=1,~x~din~ [0,2pi].~vom~aplica~formulele\\sinx=\frac{2tg\frac{x}{2} }{1+tg^{2}\frac{x}{2}} ,~cosx=\frac{1-tg^{2}\frac{x}{2}}{1+tg^{2}\frac{x}{2}}.\\Pentru~simplitate~notam~tg\frac{x}{2}=t.~obtinem~ecuatia:\\\sqrt{3}*\frac{2t}{1+t^{2}} -\frac{1-t^{2}}{1+t^{2}}=1 |*(1+t^{2}),~obtinem~2\sqrt{3}t-1+t^{2}=1+t^{2}.~2\sqrt{3}t=2,~ \sqrt{3}t=1.~t=\frac{1}{\sqrt{3}},~deci~tg\frac{x}{2}=\frac{1}{\sqrt{3}},~\frac{x}{2}=arctg\frac{1}{\sqrt{3}}+\pi  k,~k~apartine~Z.$

$\frac{x}{2}=\frac{\pi}{6}+\pi k~|*2,~x=\frac{\pi }{3}+2\pi k.\\pentru~k=0,~x=\frac{\pi }{3};\\pentru~alte~valori~ale~lui~k,~x~nu~apartine~ [0,2pi]$