Đề bài
Bài 1: Giải các phương trình sau:
a] \[2\cos [2x - \dfrac{\pi }{5}] = 1\]
b] \[\sin \left[ {x - \dfrac{\pi }{3}} \right] = \sin \left[ {2x + \dfrac{\pi }{6}} \right]\]
c] \[\sin 3x + \sin 5x = 0\]
d] \[3\tan 4x - 2\cot 4x + 1 = 0\]
Bài 2: Tìm \[x \in {\rm{[}}0;14]\] nghiệm đúng phương trình:
\[\cos 3x - 4\cos 2x + 3\cos x - 4 = 0\]
Lời giải chi tiết
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:
\[\begin{array}{l}\cos 3x - 4\cos 2x + 3\cos x - 4 = 0\\ \Leftrightarrow 4{\cos ^3}x - 3\cos x - 4[2{\cos ^2}x - 1] + 3\cos x - 4 = 0\\\Leftrightarrow 4{\cos ^3}x - 3\cos x - 8{\cos ^2}x + 4 + 3\cos x - 4 = 0 \\\Leftrightarrow 4{\cos ^3}x - 8{\cos ^2}x = 0\\\Leftrightarrow {\cos ^3}x - 2\cos {}^2x = 0 \;\; [1]\end{array}\]
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 \[x = \dfrac{\pi }{2};x = \dfrac{{3\pi }}{2};x = \dfrac{{5\pi }}{2};x = \dfrac{{7\pi }}{2}\]