giải pt: `sin4x=sin(x+π/3)`

\begin{array}{c} \color{#db1616}{\texttt{#Khoadang09}} \end{array}

Đáp án:

 `S={pi/9+k2pi;\ (2pi)/15+k2pi,\ k in ZZ}`

Giải thích các bước giải:

 `@)\ `Công thức :

`+)\ sinu=sinv`

\(\Leftrightarrow\left[ \begin{array}{l} u=v+k2\pi \\ u=\pi-v+k2\pi \end{array} \right.,\ (k \in \mathbb{Z})\)

`@)\ sin4x=sin(x+pi/3)`

\(\Leftrightarrow\left[ \begin{array}{l} 4x=x+\dfrac\pi3+k2\pi \\ 4x=\pi-x-\dfrac\pi3+k2\pi \end{array} \right.,\ (k \in \mathbb{Z})\)

\(\Leftrightarrow\left[ \begin{array}{l} 3x=\dfrac\pi5+k2\pi \\ 5x=\dfrac{2\pi}3+k2\pi \end{array} \right.,\ (k \in \mathbb{Z})\)

\(\Leftrightarrow\left[ \begin{array}{l} x=\dfrac\pi9+\dfrac{k2\pi}3 \\ x=\dfrac{2\pi}{15}+\dfrac{k2\pi}5 \end{array} \right.,\ (k \in \mathbb{Z})\)

`S={pi/9+k2pi;\ (2pi)/15+k2pi,\ k in ZZ}`