Giải phương trình: Cos(2x-Pi:2)+3sin(x-pi:4)+1=0 09/08/2021 Bởi Amaya Giải phương trình: Cos(2x-Pi:2)+3sin(x-pi:4)+1=0
Đáp án: $\left[\begin{array}{l}x = \dfrac{\pi}{12} + k2\pi\\x= \dfrac{17\pi}{12} + k2\pi \end{array}\right.\quad (k \in \Bbb Z)$ Giải thích các bước giải: $\begin{array}{l}\cos\left(2x-\dfrac{\pi}{2}\right)+3\sin\left(x-\dfrac{\pi}{4}\right)+1=0\\ \Leftrightarrow \cos\left(\dfrac{\pi}{2}-2x\right)+3\sin x\cos\dfrac{\pi}{4} – 3\cos x\sin\dfrac{\pi}{4} + 1=0\\ \Leftrightarrow \sin2x +\dfrac{3\sqrt2}{2}\sin x – \dfrac{3\sqrt2}{2}\cos x + 1=0\\ \Leftrightarrow 2\sin x\cos x +\dfrac{3\sqrt2}{2}(\sin x – \cos x) + 1=0\\ Đặt\,\,t = \sin x – \cos x, \,|t|\leq \sqrt2\\ \Rightarrow t^2 = \sin^2x + \cos^2x – 2\sin x\cos x\\ \Rightarrow 2\sin x\cos x = 1 – t^2\\ \text{Phương trình trở thành:}\\ 1 – t^2 +\dfrac{3\sqrt2}{2}t + 1=0\\ \Leftrightarrow 2t^2 – 3t\sqrt2 – 4= 0\\ \Leftrightarrow \left[\begin{array}{l}t = -\dfrac{\sqrt2}{2}\\t = 2\sqrt2\quad (loại)\end{array}\right.\\ \Leftrightarrow \sin x – \cos x = -\dfrac{\sqrt2}{2}\\ \Leftrightarrow \sqrt2\sin\left(x – \dfrac{\pi}{4}\right) = -\dfrac{\sqrt2}{2}\\ \Leftrightarrow \sin\left(x – \dfrac{\pi}{4}\right) = -\dfrac{1}{2}\\ \Leftrightarrow \left[\begin{array}{l}x – \dfrac{\pi}{4} = -\dfrac{\pi}{6} + k2\pi\\x – \dfrac{\pi}{4} = \dfrac{7\pi}{6} + k2\pi \end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k2\pi\\x= \dfrac{17\pi}{12} + k2\pi \end{array}\right.\quad (k \in \Bbb Z)\end{array}$ Bình luận
Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{12} + k2\pi\\x= \dfrac{17\pi}{12} + k2\pi \end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\cos\left(2x-\dfrac{\pi}{2}\right)+3\sin\left(x-\dfrac{\pi}{4}\right)+1=0\\ \Leftrightarrow \cos\left(\dfrac{\pi}{2}-2x\right)+3\sin x\cos\dfrac{\pi}{4} – 3\cos x\sin\dfrac{\pi}{4} + 1=0\\ \Leftrightarrow \sin2x +\dfrac{3\sqrt2}{2}\sin x – \dfrac{3\sqrt2}{2}\cos x + 1=0\\ \Leftrightarrow 2\sin x\cos x +\dfrac{3\sqrt2}{2}(\sin x – \cos x) + 1=0\\ Đặt\,\,t = \sin x – \cos x, \,|t|\leq \sqrt2\\ \Rightarrow t^2 = \sin^2x + \cos^2x – 2\sin x\cos x\\ \Rightarrow 2\sin x\cos x = 1 – t^2\\ \text{Phương trình trở thành:}\\ 1 – t^2 +\dfrac{3\sqrt2}{2}t + 1=0\\ \Leftrightarrow 2t^2 – 3t\sqrt2 – 4= 0\\ \Leftrightarrow \left[\begin{array}{l}t = -\dfrac{\sqrt2}{2}\\t = 2\sqrt2\quad (loại)\end{array}\right.\\ \Leftrightarrow \sin x – \cos x = -\dfrac{\sqrt2}{2}\\ \Leftrightarrow \sqrt2\sin\left(x – \dfrac{\pi}{4}\right) = -\dfrac{\sqrt2}{2}\\ \Leftrightarrow \sin\left(x – \dfrac{\pi}{4}\right) = -\dfrac{1}{2}\\ \Leftrightarrow \left[\begin{array}{l}x – \dfrac{\pi}{4} = -\dfrac{\pi}{6} + k2\pi\\x – \dfrac{\pi}{4} = \dfrac{7\pi}{6} + k2\pi \end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k2\pi\\x= \dfrac{17\pi}{12} + k2\pi \end{array}\right.\quad (k \in \Bbb Z)\end{array}$
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