Question

Giải phương trình sau:

\(\siṇ̣\left(4x+\dfrac{\pi}{3}\right)=-cos\left(2x+\dfrac{\pi}{3}\right)\)

Answer 1

`sin(4x+\pi/3)=-cos(2x+\pi/3)`

`<=>sin(4x+\pi/3)=cos([2\pi]/3-2x)`

`<=>sin(4x+\pi/3)=sin([-\pi]/6+2x)`

`<=>` $\left[\begin{matrix} 4x+\pi/3=[-\pi]/6+2x+k2\pi\\ 4x+\pi/3=[7\pi]/6-2x+k2\pi\end{matrix}\right.$      `(k in ZZ)`

`<=>`$\left[\begin{matrix} x=[-\pi]/4+k\pi\\ x=[5\pi]/12+[k\pi]/3\end{matrix}\right.$      `(k in ZZ)`

Vậy `S={[-\pi]/4+k\pi;[5\pi]/12+[k\pi]/3|k in ZZ}`