Bài 1:
\(\begin{array}{l}a) 2\cos (2x - \dfrac{\pi }{5}) = 1\\ \Leftrightarrow \cos (2x - \dfrac{\pi }{5}) = \dfrac{1}{2}\\ \Leftrightarrow \cos (2x - \dfrac{\pi }{5}) = \cos \dfrac{\pi }{3}\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x - \dfrac{\pi }{5} = \dfrac{\pi }{3} + k2\pi }\\{2x - \dfrac{\pi }{5} = \dfrac{{ - \pi }}{3} + k2\pi }\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{{4\pi }}{{15}} + k\pi }\\{x = \dfrac{{ - \pi }}{{15}} + k\pi }\end{array}} \right.\end{array}\)
\(b) \; \sin \left( {x - \dfrac{\pi }{3}} \right) = \sin \left( {2x + \dfrac{\pi }{6}} \right) \\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x + \dfrac{\pi }{6} = x - \dfrac{\pi }{3} + k2\pi }\\{2x + \dfrac{\pi }{6} = \pi - x + \dfrac{\pi }{3} + k2\pi }\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{{ - \pi }}{2} + k2\pi }\\{x = \dfrac{{7\pi }}{{18}} + k\dfrac{{2\pi }}{3}}\end{array}} \right.\)
\(\begin{array}{l}c)\; \sin 3x + \sin 5x = 0 \\\Leftrightarrow \sin 5x = - \sin 3x \\ \Leftrightarrow \sin 5x = \sin ( - 3x)\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{5x = - 3x + k2\pi }\\{5x = \pi + 3x + k2\pi }\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = k\dfrac{\pi }{4}}\\{x = \dfrac{\pi }{2} + k\pi }\end{array}} \right.\end{array}\)
\(d)\; 3\tan 4x - 2\cot 4x + 1 = 0\,\,\,\,(1)\)
ĐK: \(\left\{ {\begin{array}{*{20}{c}}{\sin 4x \ne 0}\\{\cos 4x \ne 0}\end{array}} \right. \)
\(\Leftrightarrow \sin 8x \ne 0 \Leftrightarrow x \ne k\dfrac{\pi }{8}\)
Đặt \(\tan 4x = t(t \ne 0) \Rightarrow \cot 4x = \dfrac{1}{t}\)
Khi đó (1) trở thành: \(3t - \dfrac{2}{t} + 1 = 0\)
\(\Leftrightarrow 3{t^2} + t - 2 = 0 \)
\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{t = - 1\,\,(TM)}\\{t = \dfrac{2}{3}\,(TM)}\end{array}} \right.\)
Với \(t = - 1 \Rightarrow \tan 4x = - 1\)
\(\Leftrightarrow \tan 4x = \tan \left( {\dfrac{{ - \pi }}{4}} \right)\)
\(\Leftrightarrow 4x = \dfrac{{ - \pi }}{4} + k\pi \)
\(\Leftrightarrow x = \dfrac{{ - \pi }}{{16}} + k\dfrac{\pi }{4}\,(TM)\)
Với \(t = \dfrac{2}{3} \Rightarrow \tan 4x = \dfrac{2}{3} \)
\(\Leftrightarrow x = \dfrac{1}{4}\arctan \dfrac{2}{3} + k\dfrac{\pi }{4}\,(TM)\)
Bài 2:
\(\begin{array}{l}\cos 3x - 4\cos 2x + 3\cos x - 4 = 0\\ \Leftrightarrow 4{\cos ^3}x - 3\cos x - 4(2{\cos ^2}x - 1) + 3\cos x - 4 = 0\\ \Leftrightarrow {\cos ^3}x - 2\cos {}^2x = 0 \;\;\;\; (1)\end{array}\)
Đặt \(\,\cos x = t\,(\left| t \right| \le 1)\)
Khi đó (1) trở thành \({t^3} - 2{t^2} = 0\)
\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{t = 0\,(TM)}\\{t = 2\,(KTM)}\end{array}} \right.\)
Với \(t = 0 \Rightarrow \cos x = 0 \)
\(\Leftrightarrow x = \dfrac{\pi }{2} + k\pi \,(k \in \mathbb{Z})\)
Mà \(x \in {\rm{[}}0;14] \Rightarrow 0 \le \dfrac{\pi }{2} + k\pi \le 14 \)\(\,\Leftrightarrow \dfrac{{ - 1}}{2} \le k \le \dfrac{{14}}{\pi } - \dfrac{1}{2}\)
Do \(k \in \mathbb{Z} \Rightarrow k \in \left\{ {0;1;2;3} \right\}\)
Vậy \(x = \dfrac{\pi }{2};x = \dfrac{{3\pi }}{2};x = \dfrac{{5\pi }}{2};x = \dfrac{{7\pi }}{2}\)
