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 \(x = π + k \frac{4\pi }{3}(k ∈ \mathbb Z)\)