Question

Am nevoie de ajutor la acest exercitiu !!!

Answer 1

Răspuns:

Aplici formulele de la final

Answer 2

$\it cos\Big(\dfrac{\pi}{3}-2x\Big)=cos\Big(2x-\dfrac{\pi}{3}\Big)\\ \\ \\ Ecua\c{\it t}ia \ este\ echivalent\breve a\ cu:\\ \\ \\ sin\Big(2x-\dfrac{\pi}{3}\Big)cosx-sinxcos\Big(2x-\dfrac{\pi}{3}\Big)=0\ \Rightarrow\ sin\Big(2x-\dfrac{\pi}{3}-x\Big)=0 \Rightarrow\\ \\ \\ \Rightarrow sin\Big(x-\dfrac{\pi}{3}\Big)=0 \ \ \ \ \ (1)\\ \\ \\ x\in(0,\ \pi)\Big|_{-\frac{\pi}{3}} \Rightarrow x-\dfrac{\pi}{3}\in\Big(-\dfrac{\pi}{3},\ \dfrac{2\pi}{3}\Big)\ \ \ \ \ (2)$

$\it (1),\ (2) \Rightarrow x-\dfrac{\pi}{3}=0 \Rightarrow x=\dfrac{\pi}{3}$