Giúp em tìm min max bài toán này với ạ Sin(x+pi/7)+sin(x-pi/7)

Đáp án:

Giải thích các bước giải:

$y = \sin \left( {x + \dfrac{\pi }{7}} \right) + \sin \left( {x - \dfrac{\pi }{7}} \right)$

Ta có: $y = \sin \left( {x + \dfrac{\pi }{7}} \right) + \sin \left( {x - \dfrac{\pi }{7}} \right)$$= 2\sin x\cos \dfrac{\pi }{7}$

Mà $- 1 \le \sin x \le 1$ nên $- 2\cos \dfrac{\pi }{7} \le 2\cos \dfrac{\pi }{7}\sin x \le 2\cos \dfrac{\pi }{7}$

Do đó $- 2\cos \dfrac{\pi }{7} \le y \le 2\cos \dfrac{\pi }{7}$.

Vậy $\max y = 2\cos \dfrac{\pi }{7}$ khi $\sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi$.

$\min y = - 2\cos \dfrac{\pi }{7}$ khi $\sin x = - 1 \Leftrightarrow x = - \dfrac{\pi }{2} + k2\pi$.

$y' = cos(x+ \pi/7) + cos(x-\pi/7)$.

y'=0 <-> $cos(x+\pi/7) = -cos(x-\pi/7)$

<-> $cos(x+\pi/7) = cos(\pi - x + \pi/7)$

<->$x + \pi/7 = 8\pi/7 -x + 2k pi$ hoac $x + \pi/7 = -8\pi/7 + x + 2k\pi$.

<->$x = \pi/2 + k\pi$

Vay y(\pi/2 + k\pi) = sin((9\pi)/14) + sin((5\pi)/14) neu k chan va bang -sin((9\pi)/14) - sin((5\pi)/14) neu k le.