[tex]\text{Ca sa intelegi mai bine :}\\
\sin x=a\text{ are solutia generala}\ \boxed{x=(-1)^k\cdot\arcsin a+k\cdot \pi ,k\in \mathbb{Z}}\\
\text{In cazul de fata:}\\
2x+\dfrac{\pi}{4}=(-1)^k\cdot \arcsin \dfrac{\sqrt{2}}{2}+k\cdot \pi,k\in \mathbb{Z}\\
2x+\dfrac{\pi}{4}=(-1)^k\cdot \dfrac{\pi}{4}+k\cdot \pi\\
\text{Ramane sa ii dam valori lui k:}\\
k=0\Rightarrow 2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}\\
x=0(\text{ asta este o solutie.})\\
\\
k=1\Rightarrow 2x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+\pi\\
x=\dfrac{\pi}{4}\\
[/tex]
[tex]k=2\Rightarrow 2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+2\cdot \pi\\
x=\pi\\
\\
k=3\Rightarrow 2x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+3\cdot \pi\\
x=\dfrac{5\pi}{4}\\
\\
k=4\Rightarrow 2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+4\cdot \pi\\
x=2\pi\notin [0,2\pi)\\
\text{Nu mai are rost sa continuam,deoarece toate solutiile vor fi mai mari}\\ \text{decat}\ 2\pi.\\
\text{Ramane sa ii dam lui k si valori negative.Te las pe tine sa faci asta.}
[/tex]
