Nội dung bài giảng
Bài 7. Giải phương trình \(f'(x) = 0\), biết rằng:
a) \(f(x) = 3\cos x + 4\sin x + 5x\);
b) \(f(x) = 1 - \sin(π + x) + 2\cos \left ( \frac{2\pi +x}{2} \right )\).
Lời giải:
a) \(f'(x) = - 3\sin x + 4\cos x + 5\). Do đó
\(f'(x) = 0 \Leftrightarrow - 3\sin x + 4\cos x + 5 = 0 \Leftrightarrow3 \sin x - 4\cos x = 5\)
\(\Leftrightarrow \frac{3}{5}\sin x - \frac{4}{5}\ cos x = 1\). (1)
Đặt \(\cos φ = \frac{3}{5}\), \(\left(φ ∈ \left ( 0;\frac{\pi }{2} \right )\right ) \Rightarrow \sin φ = \frac{4}{5}\), ta có:
(1) \(\Leftrightarrow \sin x.\cos φ - \cos x.\sin φ = 1 \Leftrightarrow \sin(x - φ) = 1\)
\(\Leftrightarrow x - φ = \frac{\pi }{2} + k2π \Leftrightarrow x = φ + \frac{\pi }{2} + k2π, k ∈ \mathbb Z\).
b) \(f'(x) = - \cos(π + x) - \sin \left (\pi + \frac{x}{2} \right ) = \cos x + \sin \frac{x }{2}\)
\(f'(x) = 0 \Leftrightarrow \cos x + \sin \frac{x }{2} = 0 \Leftrightarrow \sin \frac{x }{2} = - cosx\)
\(\Leftrightarrow sin \frac{x }{2} = sin \left (x-\frac{\pi}{2}\right )\)
\(\Leftrightarrow \frac{x }{2}= x-\frac{\pi}{2}+ k2π\) hoặc \( \frac{x }{2} = π - x+\frac{\pi}{2}+ k2π\)
\(\Leftrightarrow x = π - k4π\) hoặc \(x = π + k \frac{4\pi }{3}\), \((k ∈ \mathbb Z)\).