a) \(f'(x) = - 3\sin x + 4\cos x + 5\). Do đó
\(f'(x) = 0 \Leftrightarrow - 3\sin x + 4\cos x + 5 = 0\)
\(\Leftrightarrow3 \sin x - 4\cos x = 5\)
\(\Leftrightarrow \dfrac{3}{5}\sin x - \dfrac{4}{5}\ cos x = 1\). (1)
Đặt \(\cos φ = \dfrac{3}{5}\), \(\left(φ ∈ \left ( 0;\dfrac{\pi }{2} \right )\right ) \Rightarrow \sin φ = \dfrac{4}{5}\), ta có:
(1) \(\Leftrightarrow \sin x.\cos φ - \cos x.\sin φ = 1 \Leftrightarrow \sin(x - φ) = 1\)
\(\Leftrightarrow x - φ = \dfrac{\pi }{2} + k2π \Leftrightarrow x = φ + \dfrac{\pi }{2} + k2π, k ∈ \mathbb Z\)
b) \(f'(x) = - \cos(π + x) - \sin \left (\pi + \dfrac{x}{2} \right ) = \cos x + \sin \dfrac{x }{2}\)
\(f'(x) = 0 \Leftrightarrow \cos x + \sin \dfrac{x }{2} = 0 \Leftrightarrow \sin \dfrac{x }{2} = - cosx\)
\(\Leftrightarrow sin \dfrac{x }{2} = sin \left (x-\dfrac{\pi}{2}\right )\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\dfrac{x}{2} = x - \dfrac{\pi }{2} + k2\pi \\\dfrac{x}{2} = \pi - x + \dfrac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{4} - k4\pi \\x = \pi + \dfrac{{4k\pi }}{3}\end{array} \right.\,\,\left( {k \in Z} \right)\end{array}\)
