Giải phương trình a) 2sin3x-1=0 b) sin^2 3x-4cos3x +4=0 c) sin4x=cos 4x d) cos ^2 3x -4 cos 3x+3=0 05/12/2021 Bởi Aaliyah Giải phương trình a) 2sin3x-1=0 b) sin^2 3x-4cos3x +4=0 c) sin4x=cos 4x d) cos ^2 3x -4 cos 3x+3=0
Đáp án: $\begin{array}{l}a)\quad \left[\begin{array}{l}x = \dfrac{\pi}{18} + k\dfrac{2\pi}{3}\\x = \dfrac{5\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)\\b)\quad x = k\dfrac{2\pi}{3}\quad (k\in\Bbb Z)\\c)\quad x = \dfrac{\pi}{16} + k\dfrac{\pi}{4}\quad (k\in\Bbb Z)\\d)\quad x=k\dfrac{2\pi}{3}\quad (k\in\Bbb Z)\end{array}$ Giải thích các bước giải: $\begin{array}{l}a)\quad 2\sin3x – 1 =0\\ \Leftrightarrow \sin3x = \dfrac12\\ \Leftrightarrow \sin3x = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}3x = \dfrac{\pi}{6} + k2\pi\\3x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{18} + k\dfrac{2\pi}{3}\\x = \dfrac{5\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)\\ b)\quad \sin^23x – 4\cos3x + 4 = 0\\ \Leftrightarrow 1 – \cos^23x – 4\cos3x + 4 = 0\\ \Leftrightarrow \cos^23x + 4\cos3x – 5 = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos3x = 1\\\cos3x = -5\quad (loại)\end{array}\right.\\ \Leftrightarrow 3x = k2\pi\\ \Leftrightarrow x = k\dfrac{2\pi}{3}\quad (k\in\Bbb Z)\\ c)\quad \sin4x = \cos4x\\ \Leftrightarrow \sin4x – \cos4x = 0\\ \Leftrightarrow \sin\left(4x – \dfrac{\pi}{4}\right) = 0\\ \Leftrightarrow 4x – \dfrac{\pi}{4} = k\pi\\ \Leftrightarrow 4x = \dfrac{\pi}{4} + k\pi\\ \Leftrightarrow x = \dfrac{\pi}{16} + k\dfrac{\pi}{4}\quad (k\in\Bbb Z)\\ d)\quad \cos^23x – 4\cos3x + 3 =0\\ \Leftrightarrow \left[\begin{array}{l}\cos3x = 1\\\cos3x = 3\quad (loại)\end{array}\right.\\ \Leftrightarrow 3x = k2\pi\\ \Leftrightarrow x=k\dfrac{2\pi}{3}\quad (k\in\Bbb Z) \end{array}$ Bình luận
Đáp án:
BẠN THAM KHẢO NHA!!!
Đáp án:
$\begin{array}{l}a)\quad \left[\begin{array}{l}x = \dfrac{\pi}{18} + k\dfrac{2\pi}{3}\\x = \dfrac{5\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)\\b)\quad x = k\dfrac{2\pi}{3}\quad (k\in\Bbb Z)\\c)\quad x = \dfrac{\pi}{16} + k\dfrac{\pi}{4}\quad (k\in\Bbb Z)\\d)\quad x=k\dfrac{2\pi}{3}\quad (k\in\Bbb Z)\end{array}$
Giải thích các bước giải:
$\begin{array}{l}a)\quad 2\sin3x – 1 =0\\ \Leftrightarrow \sin3x = \dfrac12\\ \Leftrightarrow \sin3x = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}3x = \dfrac{\pi}{6} + k2\pi\\3x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{18} + k\dfrac{2\pi}{3}\\x = \dfrac{5\pi}{18} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)\\ b)\quad \sin^23x – 4\cos3x + 4 = 0\\ \Leftrightarrow 1 – \cos^23x – 4\cos3x + 4 = 0\\ \Leftrightarrow \cos^23x + 4\cos3x – 5 = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos3x = 1\\\cos3x = -5\quad (loại)\end{array}\right.\\ \Leftrightarrow 3x = k2\pi\\ \Leftrightarrow x = k\dfrac{2\pi}{3}\quad (k\in\Bbb Z)\\ c)\quad \sin4x = \cos4x\\ \Leftrightarrow \sin4x – \cos4x = 0\\ \Leftrightarrow \sin\left(4x – \dfrac{\pi}{4}\right) = 0\\ \Leftrightarrow 4x – \dfrac{\pi}{4} = k\pi\\ \Leftrightarrow 4x = \dfrac{\pi}{4} + k\pi\\ \Leftrightarrow x = \dfrac{\pi}{16} + k\dfrac{\pi}{4}\quad (k\in\Bbb Z)\\ d)\quad \cos^23x – 4\cos3x + 3 =0\\ \Leftrightarrow \left[\begin{array}{l}\cos3x = 1\\\cos3x = 3\quad (loại)\end{array}\right.\\ \Leftrightarrow 3x = k2\pi\\ \Leftrightarrow x=k\dfrac{2\pi}{3}\quad (k\in\Bbb Z) \end{array}$