Giải câu 7 bài 3: Đạo hàm của hàm số lượng giác.
a) \(f(x) = 3\cos x + 4\sin x + 5x\)
\(f'(x) = - 3\sin x + 4\cos x + 5\).
\(\Rightarrow f'(x) = 0 \Leftrightarrow - 3\sin x + 4\cos x + 5 = 0\)
\(\Leftrightarrow3 \sin x - 4\cos x = 5\)
\(\Leftrightarrow \frac{3}{5}\sin x - \frac{4}{5}\ cos x = 1\).(*)
Đặt \(\cos \alpha = \frac{3}{5},\left(\alpha ∈ \left ( 0;\frac{\pi }{2} \right )\right ) \Rightarrow \sin \alpha = \frac{4}{5}\)
Ta có:
(*)\(\Leftrightarrow \sin x.\cos \alpha - \cos x.\sin \alpha = 1\)
\(\Leftrightarrow \sin(x - \alpha ) = 1\)
\(\Leftrightarrow x - \alpha = \frac{\pi }{2} + k2π\)
\(\Leftrightarrow x = \alpha + \frac{\pi }{2} + k2π, k ∈ \mathbb Z\).
Vậy \(x = \alpha + \frac{\pi }{2} + k2π, k ∈ \mathbb Z\)
b) \(f(x) = 1 - \sin(π + x) + 2\cos \left ( \frac{2\pi +x}{2} \right )\)
\(f'(x) = - \cos(π + x) - \sin \left (\pi + \frac{x}{2} \right ) = \cos x + \sin \frac{x }{2}\)
\(f'(x) = 0 \Leftrightarrow \cos x + \sin \frac{x }{2} = 0 \)
\(\Leftrightarrow \sin \frac{x }{2} = - cosx\)
\(\Leftrightarrow sin \frac{x }{2} = sin \left (x-\frac{\pi}{2}\right )\)
\(\Leftrightarrow \left[ \matrix{\frac{x }{2}= x-\frac{\pi}{2}+ k2π \hfill \cr \frac{x }{2} = π - x+\frac{\pi}{2}+ k2π \hfill \cr} \right.\)
\(\Leftrightarrow \left[ \matrix{x = π - k4π \hfill \cr x = π + k \frac{4\pi }{3} \hfill \cr} \right.(k ∈ \mathbb Z)\)
Vậy \(x = π - k4π\)hoặc \(x = π + k \frac{4\pi }{3}(k ∈ \mathbb Z)\)