Giải câu 7 bài 3: Đạo hàm của hàm số lượng giác

a) (f(x) = 3cos x + 4sin x + 5x)

(f'(x) = – 3sin x + 4cos x + 5).

(Rightarrow f'(x) = 0 Leftrightarrow – 3sin x + 4cos x + 5 = 0)

(Leftrightarrow3 sin x – 4cos 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) + 2cos left ( frac{2pi +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{4pi }{3} hfill cr} right.(k ∈ mathbb Z))

Vậy (x = π – k4π)hoặc (x = π + k frac{4pi }{3}(k ∈ mathbb Z))