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Bài 7. Giải phương trình \(f'(x) = 0\), biết rằng:
a) \(f(x) = 3\cos x + 4\sin x + 5x\);
b) \(f(x) = 1 – \sin(π + x) + 2\cos \left ( \frac{2\pi +x}{2} \right )\).
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) \(\Leftrightarrow \sin x.\cos φ – \cos x.\sin φ = 1 \Leftrightarrow \sin(x – φ) = 1\)
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\(\Leftrightarrow x – φ = \frac{\pi }{2} + k2π \Leftrightarrow x = φ + \frac{\pi }{2} + k2π, k ∈ \mathbb Z\).
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 \(x = π + k \frac{4\pi }{3}\), \((k ∈ \mathbb Z)\).