Salut,
Ai obținut o ecuație trigonometrică clasică. Rezolvarea ei este așa:
[tex]sin\left(\dfrac{\pi}5+a\right)=0\Rightarrow \dfrac{\pi}5+a=(-1)^k\cdot arcsin 0+k\cdot\pi\Rightarrow\\\\\Rightarrow \dfrac{\pi}5+a=(-1)^k\cdot 0+k\cdot\pi\Rightarrow \dfrac{\pi}5+a=k\cdot\pi\Rightarrow a=k\cdot\pi-\dfrac{\pi}5=\pi\cdot\left(k-\dfrac{1}5\right).\\\\Cum\ a\in(0,\pi)\Rightarrow 0<\pi\cdot\left(k-\dfrac{1}5\right)<\pi\ \Bigg{|}:\pi\Rightarrow 0<k-\dfrac{1}5<1\ \Bigg{|}+\dfrac{1}5\Rightarrow\\\\\Rightarrow\dfrac{1}5<k<\dfrac{6}5.\ Dar\ k\in\mathbb{Z},\ deci\ k=1\ care\ este\ singura\ valoare\ \hat{\i}ntreag\breve{a}\ din\left(\dfrac{1}5,\dfrac{6}5\right).\\\\La\ final,\ a=k\cdot\pi-\dfrac{\pi}5=\pi-\dfrac{\pi}5=\dfrac{4\pi}5.[/tex]
Ai înțeles rezolvarea ?
Green eyes.
