giúp tui vs 2sinx -cosx =2 Cos3x+sin3x=1 √3 sin2x -cosx =1 13/07/2021 Bởi Ruby giúp tui vs 2sinx -cosx =2 Cos3x+sin3x=1 √3 sin2x -cosx =1
Đáp án: $\begin{array}{l}a)\, \left[\begin{array}{l}x = \dfrac{\pi}{2}+ k2\pi\\x = \dfrac{\pi}{2} + 2\arccos\dfrac{2}{\sqrt5} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\b)\,\left[\begin{array}{l}x = k2\pi\\x =\dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\c)\,\left[\begin{array}{l}x = \dfrac{\pi}{6} + k\pi\\x = \dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\end{array}$ Giải thích các bước giải: $\begin{array}{l}a)\, 2\sin x – \cos x = 2\\ \Leftrightarrow \dfrac{2}{\sqrt5}\sin x – \dfrac{1}{\sqrt5}\cos x = \dfrac{2}{\sqrt5}\\ Do\,\,\left(\dfrac{2}{\sqrt5}\right)^2 + \left(\dfrac{1}{\sqrt5}\right)^2 = 1\\ Đặt\,\begin{cases}\cos\alpha = \dfrac{2}{\sqrt5}\\\sin\alpha = \dfrac{1}{\sqrt5}\end{cases}\Rightarrow \alpha = \arccos\dfrac{2}{\sqrt5}\\ \text{Phương trình trở thành:}\\ \sin x\cos\alpha – \cos x\sin\alpha = \cos\alpha\\ \Leftrightarrow \sin(x – \alpha) = \sin\left(\dfrac{\pi}{2} – \alpha\right)\\ \Leftrightarrow \left[\begin{array}{l}x – \alpha = \dfrac{\pi}{2} – \alpha + k2\pi\\x – \alpha = \dfrac{\pi}{2} + \alpha + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2}+ k2\pi\\x = \dfrac{\pi}{2} + 2\alpha + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2}+ k2\pi\\x = \dfrac{\pi}{2} + 2\arccos\dfrac{2}{\sqrt5} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ b)\,\cos3x + \sin3x = 1\\ \Leftrightarrow \sqrt2\sin\left(3x + \dfrac{\pi}{4}\right) = 1\\ \Leftrightarrow \sin\left(3x + \dfrac{\pi}{4}\right) = \dfrac{\sqrt2}{2}\\ \Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = \dfrac{\pi}{4} + k2\pi\\x + \dfrac{\pi}{4} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k2\pi\\x =\dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ c)\, \sqrt3\sin2x – \cos2x = 1\\ \Leftrightarrow \dfrac{\sqrt3}{2}\sin2x – \dfrac{1}{2}\cos2x = \dfrac{1}{2}\\ \Leftrightarrow \sin\left(2x -\dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}2x – \dfrac{\pi}{6} = \dfrac{\pi}{6} + k2\pi\\2x – \dfrac{\pi}{6} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{3} + k2\pi\\2x = \pi + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} + k\pi\\x = \dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$ Bình luận
Đáp án:
$\begin{array}{l}a)\, \left[\begin{array}{l}x = \dfrac{\pi}{2}+ k2\pi\\x = \dfrac{\pi}{2} + 2\arccos\dfrac{2}{\sqrt5} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\b)\,\left[\begin{array}{l}x = k2\pi\\x =\dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\c)\,\left[\begin{array}{l}x = \dfrac{\pi}{6} + k\pi\\x = \dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\end{array}$
Giải thích các bước giải:
$\begin{array}{l}a)\, 2\sin x – \cos x = 2\\ \Leftrightarrow \dfrac{2}{\sqrt5}\sin x – \dfrac{1}{\sqrt5}\cos x = \dfrac{2}{\sqrt5}\\ Do\,\,\left(\dfrac{2}{\sqrt5}\right)^2 + \left(\dfrac{1}{\sqrt5}\right)^2 = 1\\ Đặt\,\begin{cases}\cos\alpha = \dfrac{2}{\sqrt5}\\\sin\alpha = \dfrac{1}{\sqrt5}\end{cases}\Rightarrow \alpha = \arccos\dfrac{2}{\sqrt5}\\ \text{Phương trình trở thành:}\\ \sin x\cos\alpha – \cos x\sin\alpha = \cos\alpha\\ \Leftrightarrow \sin(x – \alpha) = \sin\left(\dfrac{\pi}{2} – \alpha\right)\\ \Leftrightarrow \left[\begin{array}{l}x – \alpha = \dfrac{\pi}{2} – \alpha + k2\pi\\x – \alpha = \dfrac{\pi}{2} + \alpha + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2}+ k2\pi\\x = \dfrac{\pi}{2} + 2\alpha + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2}+ k2\pi\\x = \dfrac{\pi}{2} + 2\arccos\dfrac{2}{\sqrt5} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ b)\,\cos3x + \sin3x = 1\\ \Leftrightarrow \sqrt2\sin\left(3x + \dfrac{\pi}{4}\right) = 1\\ \Leftrightarrow \sin\left(3x + \dfrac{\pi}{4}\right) = \dfrac{\sqrt2}{2}\\ \Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = \dfrac{\pi}{4} + k2\pi\\x + \dfrac{\pi}{4} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k2\pi\\x =\dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ c)\, \sqrt3\sin2x – \cos2x = 1\\ \Leftrightarrow \dfrac{\sqrt3}{2}\sin2x – \dfrac{1}{2}\cos2x = \dfrac{1}{2}\\ \Leftrightarrow \sin\left(2x -\dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}2x – \dfrac{\pi}{6} = \dfrac{\pi}{6} + k2\pi\\2x – \dfrac{\pi}{6} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{3} + k2\pi\\2x = \pi + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} + k\pi\\x = \dfrac{\pi}{2} + k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$