Làm hộ e câu b với bài 2 kia với ạ Thanks

Đáp án:

$b)\,\,\left[\begin{array}{l}x = – \dfrac{\pi}{2} + k2\pi\\x = – \dfrac{3\pi}{4} +k2\pi\end{array}\right.\quad (k \in \Bbb Z)$

$2) \,\,\begin{cases}\min y = 4 \Leftrightarrow  x = \dfrac{\pi}{2} + k\pi\\\max y = 5 \Leftrightarrow x = k\pi \quad (k \in \Bbb Z)\end{cases}$

Giải thích các bước giải:

$\begin{array}{l}b)\,\,\dfrac{\cos^3x – \cos^2x}{\sin x + \cos x} = 2(1 + \sin x) \qquad (*)\\ ĐKXĐ:\, \sin x + \cos x \ne 0 \Leftrightarrow x \ne -\dfrac{\pi}{4} + n\pi \quad (n \in \Bbb Z)\\ (*) \Leftrightarrow \cos^2x(\cos x – 1) = 2(1 + \sin x)(\sin x + \cos x)\\ \Leftrightarrow (1 + \sin x)(1 – \sin x)(\cos x – 1) – 2(1 + \sin x)(\sin x + \cos x) = 0\\ \Leftrightarrow (1 + \sin x)(\sin x + \cos x – \sin x\cos x – 1 – 2(\sin x + \cos x)) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = -1\\\sin x + \cos x + \sin x\cos x + 1 = 0 \quad (**)\end{array}\right.\\ Đặt\,\,t = \sin x + \cos x\qquad (|t| \leq \sqrt2)\\ \Rightarrow t^2 = 1 + 2\sin x\cos x\\ \Rightarrow \dfrac{t^2 – 1}{2} = \sin x\cos x\\ (**) \,\,trở\,\,thành:\\ t + \dfrac{t^2 – 1}{2} + 1 = 0\\ \Leftrightarrow t^2 + 2t + 1 = 0\\ \Leftrightarrow (t + 1)^2 = 0\\ \Leftrightarrow t = -1\\ Ta\,\,được:\\ (*) \Leftrightarrow \left[\begin{array}{l}\sin x = -1\\\sin x + \cos x = -1\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{2} + k2\pi\\\sin\left(x + \dfrac{\pi}{4}\right)= -1\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{2} + k2\pi\\x = – \dfrac{3\pi}{4} +k2\pi\end{array}\right.\quad (k \in \Bbb Z)\end{array}$

$2)\,\, y = 3 + \sqrt{4 – 3\sin^2x}$

Ta có:

$0 \leq \sin^2x \leq 1$

$\Leftrightarrow -3 \leq -3\sin^2x \leq 0$

$\Leftrightarrow 1 \leq 4 – 3\sin^2x \leq 4$

$\Leftrightarrow 1 \leq \sqrt{4 – 3\sin^2x} \leq 2$

$\Leftrightarrow 4 \leq 3 + \sqrt{4 – 3\sin^2x} \leq 5$

Hay $4 \leq y \leq 5$

Vậy $\min y = 4 \Leftrightarrow \sin^2x = 1 \Leftrightarrow \sin x = \pm 1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi$

$\max y = 5 \Leftrightarrow \sin x = 0 \Leftrightarrow x = k\pi \quad (k \in \Bbb Z)$