Giải phương trình sin4x + cos3x -cosx =0

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`sin 4x + cos 3x - cos x = 0` $\\$ `<=> 2sin2x.cos2x - 2sin2x.sinx = 0` $\\$ `<=> 2sin2x(cos2x - sinx) = 0` $\\$ `<=>` $\left[\begin{matrix} sin 2x = 0\\ cos 2x - sin x = 0\end{matrix}\right.$ $\\$ `TH1 : sin 2x = 0` $\\$ `<=> 2x = k.\pi` $\text{(k $\in$ $\mathbb{Z}$)}$ $\\$ `<=> x = \frac{k.\pi}{2}` $\text{(k $\in$ $\mathbb{Z}$)}$ $\\$ `TH2: cos 2x - sin x = 0` $\\$ `<=> cos 2x = sin x` $\\$ `<=> cos 2x = cos (\frac{\pi}{2} - x)` $\\$ `<=>` $\left[\begin{matrix} 2x = \frac{\pi}{2} - x + k.2\pi\\ 2x = - \frac{\pi}{2} + x + k.2\pi\end{matrix}\right.$ $\text{(k $\in$ $\mathbb{Z}$)}$ $\\$ `<=>` $\left[\begin{matrix} x = \frac{\pi}{6} + \frac{k.2\pi}{3}\\ x = - \frac{\pi}{2} + k.2\pi\end{matrix}\right.$ $\text{(k $\in$ $\mathbb{Z}$)}$

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