Giải phương trình f'(x)=0, biết rằng

$$\begin{gathered} a, f\left(x\right)=3\mathrm{co}sx +4\sin x+5x \\ b, f\left(x\right)=1-\sin \left(\mathrm{\pi }+\mathrm{x}\right)+2\mathrm{co}s\left(\frac{2\mathrm{\pi }+\mathrm{x}}{2}\right) \\ Giải \\ a, Ta có f\text{'}\left(x\right)=-3\sin x +4\cos x +5 \\ f\text{'}\left(x\right)=0\iff -3\sin x +4\cos x +5=0 \\ \iff 3\sin x -4\cos x =5 \iff \frac{3}{5}\sin x-\frac{4}{5}\cos x =1 \\ \iff \cos x\cos \phi -\sin xsin\phi =-1 ( Với \cos \phi =\frac{4}{5}) \\ \iff \cos \left(x-\phi \right)=-1\iff x=\mathrm{\pi }+\mathrm{\phi }+\mathrm{k}2\mathrm{\pi }\left(\mathrm{k}\in \mathrm{\mathbb{Z}}\right) \\ \mathrm{b},\mathrm{f}\text{'}\left(\mathrm{x}\right)=-\cos \left(\mathrm{\pi }+\mathrm{x}\right)-\sin \left(\mathrm{\pi }+\frac{\mathrm{x}}{2}\right)=\mathrm{cosx} +\sin \frac{\mathrm{x}}{2} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=0\iff \sin \frac{\mathrm{x}}{2}=-\mathrm{cosx}\iff \sin \frac{\mathrm{x}}{2}=\sin \left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right) \\ \iff \left[\begin{array}{c}\frac{\mathrm{x}}{2}=\mathrm{x}-\frac{\mathrm{\pi }}{2}+\mathrm{k}2\mathrm{\pi }\\ \frac{\mathrm{x}}{2}=\mathrm{\pi }-\mathrm{x}+\frac{\mathrm{\pi }}{2}+\mathrm{k}2\mathrm{\pi }\end{array}\right] \iff \left[\begin{array}{c}\mathrm{x}=\mathrm{\pi }-\mathrm{k}4\mathrm{\pi }\\ \mathrm{x}=\mathrm{\pi }+\mathrm{k}\frac{4\mathrm{\pi }}{3}\end{array}\right] \mathrm{k}\in \mathrm{\mathbb{Z}} \iff \mathrm{x}=\mathrm{\pi }+\mathrm{k}\frac{4\mathrm{\pi }}{3};\mathrm{k}\in \mathrm{\mathbb{Z}} \end{gathered}$$