Giải phương trình f′(x) = 0, biết rằng f(x) = 3cosx + 4sinx + 5x

Giải Toán 11 Bài 3: Đạo hàm của hàm số lượng giác

Bài tập 7 trang 169 SGK Toán lớp 11 Đại số:

a) f(x) = 3cosx + 4sinx + 5x;

b) $y=1-\sin \left(\pi +x\right)+2\cos \left(\frac{2\pi +x}{2}\right)$;

Lời giải:

a) f(x) = 3cosx + 4sinx + 5x

f′(x) = –3sinx + 4cosx + 5

Do đó: f′(x) = 0

$\iff$ –3sinx + 4cosx + 5 = 0

$\iff$ 3sinx – 4cosx = 5

$\iff \frac{3}{5}\sin x-\frac{4}{5}\cos x=1$  (1)

Đặt $\cos \phi =\frac{3}{5},\left(\phi \in \left(0;\frac{\pi }{2}\right)\right)\Rightarrow \sin \phi =\frac{4}{5}$, ta có:

(1) $\iff \sin x.\cos \phi -\cos x.\sin \phi =1$

$\iff \sin (x-\phi )=1\iff x-\phi =\frac{\pi }{2}+k2\pi \iff x=\phi +\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z}$

b)$y=1-\sin \left(\pi +x\right)+2\cos \left(\frac{2\pi +x}{2}\right)$

$f^{\text{'}}\left(x\right)=(1{)}^{\text{'}}-[\sin (\pi +x){]}^{\text{'}}+2{\left.\cos \left(\pi +\frac{x}{2}\right)\right]}^{\text{'}}=-(\pi +x{)}^{\text{'}}\cos (\pi +x)+2{\left(\pi +\frac{x}{2}\right)}^{\text{'}}\cdot \left.-\sin \left(\pi +\frac{x}{2}\right)\right]=-\cos (\pi +x)+2\cdot \frac{1}{2}\cdot \left.-\sin \left(\pi +\frac{x}{2}\right)\right]f^{\text{'}}(x)=-\cos (\pi +x)-\sin \left(\pi +\frac{x}{2}\right)=\cos x+\sin \frac{x}{2}{f}^{\text{'}}(x)=0\iff \cos x+\sin \frac{x}{2}=0\iff \sin \frac{x}{2}=-\cos x\iff \sin \frac{x}{2}=\sin \left(x-\frac{\pi }{2}\right)\iff \begin{array}{l}\frac{x}{2}=x-\frac{\pi }{2}+k2\pi \\ \frac{x}{2}=\pi -x+\frac{\pi }{2}+k2\pi \end{array}\iff \begin{array}{l}-\frac{x}{2}=-\frac{\pi }{2}+k2\pi \\ \frac{3x}{2}=\frac{3\pi }{2}+k2\pi \end{array}\iff \begin{array}{l}x=\pi -k4\pi \\ x=\pi +\frac{k4\pi }{3}\end{array}\iff x=\pi +\frac{k4\pi }{3}$

Cách khác:

$f\left(x\right)=1-\sin \left(\pi +x\right)+2\cos \left(\frac{2\pi +x}{2}\right)=1+\sin x+2\cos \left(\pi +\frac{x}{2}\right)=1+\sin x-2\cos \frac{x}{2}{f}^{\text{'}}\left(x\right)={\left(1+\sin x-2\cos \frac{x}{2}\right)}^{\text{'}}=(1{)}^{\text{'}}+(\sin x{)}^{\text{'}}-2{\left(\cos \frac{x}{2}\right)}^{\text{'}}=0+\cos x-2\cdot \frac{1}{2}\left(-\sin \frac{x}{2}\right)=\cos x+\sin \frac{x}{2}{f}^{\text{'}}\left(x\right)=0\iff \cos x+\sin \frac{x}{2}=0\iff \cos x=-\sin \frac{x}{2}=-\cos \left(\frac{\pi }{2}-\frac{x}{2}\right)\iff \cos x=\cos \left(\pi -\left(\frac{\pi }{2}-\frac{x}{2}\right)\right)\iff \cos x=\cos \left(\frac{\pi }{2}+\frac{x}{2}\right)\iff \begin{array}{l}x=\frac{\pi }{2}+\frac{x}{2}+k2\pi \\ x=-\frac{\pi }{2}-\frac{x}{2}+k2\pi \end{array}\iff \begin{array}{l}\frac{3x}{2}=-\frac{\pi }{2}+k2\pi \\ \frac{x}{2}=\frac{\pi }{2}+k2\pi \end{array}\iff \begin{array}{l}x=\pi +k4\pi \\ x=-\frac{\pi }{3}+\frac{k4\pi }{3}\end{array}$