Toán Nghiệm của phương trình sin ( x + 2pi / 3 ) = cos3x 24/07/2021 By Nevaeh Nghiệm của phương trình sin ( x + 2pi / 3 ) = cos3x
$sin\Bigg(x+\dfrac{2\pi}{3}\Bigg)=cos3x$ $↔ sin\Bigg(x+\dfrac{2\pi}{3}\Bigg)=sin\Bigg(\dfrac{\pi}{2}-3x\Bigg)$ $↔ \left[ \begin{array}{l}x+\dfrac{2\pi}{3}=\dfrac{\pi}{2}-3x+k2\pi\\x+\dfrac{2\pi}{3}=\dfrac{\pi}{2}+3x+k2\pi\end{array} \right.$ $↔ \left[ \begin{array}{l}x=-\dfrac{\pi}{24}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{12}+k\pi\end{array} \right.$ $(k∈Z)$ Trả lời
Đáp án: \(\left[ \begin{array}{l}x = -\frac{π}{24} + k\frac{π}{2}\\x = \frac{π}{12} + kπ\end{array} \right.\) `(k ∈ ZZ)` Giải thích các bước giải: `sin (x + (2π)/3) = cos 3x` `<=> sin (x + (2π)/3) = sin ((π)/2 – 3x)` `<=>` \(\left[ \begin{array}{l}x + \frac{2π}{3} = \frac{π}{2} – 3x + k2π\\x + \frac{2π}{3} = \frac{π}{2} + 3x + k2π\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = -\frac{π}{24} + k\frac{π}{2}\\x = \frac{π}{12} + kπ\end{array} \right.\) `(k ∈ ZZ)` Trả lời
$sin\Bigg(x+\dfrac{2\pi}{3}\Bigg)=cos3x$
$↔ sin\Bigg(x+\dfrac{2\pi}{3}\Bigg)=sin\Bigg(\dfrac{\pi}{2}-3x\Bigg)$
$↔ \left[ \begin{array}{l}x+\dfrac{2\pi}{3}=\dfrac{\pi}{2}-3x+k2\pi\\x+\dfrac{2\pi}{3}=\dfrac{\pi}{2}+3x+k2\pi\end{array} \right.$
$↔ \left[ \begin{array}{l}x=-\dfrac{\pi}{24}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{12}+k\pi\end{array} \right.$ $(k∈Z)$
Đáp án: \(\left[ \begin{array}{l}x = -\frac{π}{24} + k\frac{π}{2}\\x = \frac{π}{12} + kπ\end{array} \right.\) `(k ∈ ZZ)`
Giải thích các bước giải:
`sin (x + (2π)/3) = cos 3x`
`<=> sin (x + (2π)/3) = sin ((π)/2 – 3x)`
`<=>` \(\left[ \begin{array}{l}x + \frac{2π}{3} = \frac{π}{2} – 3x + k2π\\x + \frac{2π}{3} = \frac{π}{2} + 3x + k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = -\frac{π}{24} + k\frac{π}{2}\\x = \frac{π}{12} + kπ\end{array} \right.\) `(k ∈ ZZ)`