Giải PT: sin3x-cos5x=0

Bạn tham khảo:

sin3x -cos5x = 0

<=> sin3x = cos5x

<=> sin3x = sin($\frac{\pi}{2}$ -5x) $\\$ <=> $\left[\begin{matrix} 3x = \frac{\pi}{2} - 5x + k.2\pi\\ 3x = \pi - \frac{\pi}{2} + 5x + k.2\pi\end{matrix}\right.$ (K $\in$ $\mathbb{Z}$) $\\$ <=> $\left[\begin{matrix} 8x = \frac{\pi}{2} + k.2\pi\\ -2x = \frac{\pi}{2} + k.2\pi\end{matrix}\right.$ (K $\in$ $\mathbb{Z}$) $\\$ <=> $\left[\begin{matrix} x = \frac{\pi}{16} + \frac{k\pi}{4}\\ x = - \frac{\pi}{4} - k.2\pi\end{matrix}\right.$ (K $\in$ $\mathbb{Z}$)

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