3sin^2 2X +4sin2x +1=0 3cos^2x +5cosx +2=0 2sin^2 (x+30°) -3sin(x+30°)-5=0 Tan^2 3X +tan3x -2=0

Đáp án:

$\begin{array}{l}a) \,\,\left[\begin{array}{l}x = – \dfrac{\pi}{4} + k\pi\\x = \dfrac{1}{2}\arcsin\left(-\dfrac{1}{3}\right) + k\pi\\x = \dfrac{\pi}{2} – \dfrac{1}{2}\arcsin\left(-\dfrac{1}{3}\right) + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\b)\,\,\left[\begin{array}{l}x = \pi + k2\pi\\x = \pm \arccos\left(-\dfrac{2}{3}\right) + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\c)\,\,x = -120^o + k.360^o\quad (k \in \Bbb Z)\\d)\,\,\left[\begin{array}{l}x = -\dfrac{\pi}{12} +k\dfrac{\pi}{3}\\x = \dfrac{1}{3}\arctan(-2) + k\dfrac{\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)\end{array}$

Giải thích các bước giải:

$\begin{array}{l}a)\,\,3\sin^22x + 4\sin2x + 1 = 0\\ \Leftrightarrow (\sin2x +1)\left(\sin2x + \dfrac{1}{3}\right) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin2x = – 1\\\sin2x = – \dfrac{1}{3}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = – \dfrac{\pi}{2} + k2\pi\\2x = \arcsin\left(-\dfrac{1}{3}\right) + k2\pi\\2x = \pi – \arcsin\left(-\dfrac{1}{3}\right) + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{4} + k\pi\\x = \dfrac{1}{2}\arcsin\left(-\dfrac{1}{3}\right) + k\pi\\x = \dfrac{\pi}{2} – \dfrac{1}{2}\arcsin\left(-\dfrac{1}{3}\right) + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ b)\,\,3\cos^2x +5\cos x +2=0\\ \Leftrightarrow (\cos x + 1)\left(\cos x + \dfrac{2}{3}\right) = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos x = – 1\\\cos x = – \dfrac{2}{3}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \pi + k2\pi\\x = \pm \arccos\left(-\dfrac{2}{3}\right) + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ c)\,\,2\sin^2(x+30^o) -3\sin(x+30^o)-5=0\\ \Leftrightarrow [\sin(x+ 30^o)+1].\left[\sin(x+ 30^o) – \dfrac{5}{2}\right] = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin(x + 30^o) = -1\\\sin x = \dfrac{5}{2} > 1 \quad (loại)\end{array}\right.\\ \Leftrightarrow x + 30^o = – 90^o +k.360^o\\ \Leftrightarrow x = -120^o + k.360^o\quad (k \in \Bbb Z)\\ d)\,\,\tan^23x +\tan3x -2=0\quad (*)\\ ĐKXĐ: \, \cos3x \ne 0 \Leftrightarrow x \ne \dfrac{\pi}{6} +n\dfrac{\pi}{3}\quad (n \in \Bbb Z)\\ (*)\Leftrightarrow (\tan3x – 1)(\tan3x + 2) =0\\ \Leftrightarrow \left[\begin{array}{l}\tan3x = -1\\\tan3x = – 2\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}3x = -\dfrac{\pi}{4} +k\pi\\3x = \arctan(-2) + k\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{12} +k\dfrac{\pi}{3}\\x = \dfrac{1}{3}\arctan(-2) + k\dfrac{\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)\\ \end{array}$