Question

Giải pt sau A) Sin 3x+1/3=0 B) sin (pi/3-x)=0

Answer 1

a: \(\Leftrightarrow\sin3x=-\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=arcsin\left(-\dfrac{1}{3}\right)+k2\Pi\\3x=\Pi-arcsin\left(-\dfrac{1}{3}\right)+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(arcsin\left(-\dfrac{1}{3}\right)+k2\Pi\right)\\x=\dfrac{1}{3}\left(\Pi-arcsin\left(-\dfrac{1}{3}\right)+k2\Pi\right)\end{matrix}\right.\)

b: \(\Leftrightarrow-x+\dfrac{\Pi}{3}=k\Pi\)

hay \(x=-k\Pi+\dfrac{\Pi}{3}\)