Đề bài
Bài 1: Giải các phương trình lượng giác sau:
a] \[\tan \left[ {2x - 1} \right] = \sqrt 3 \]
b] \[\cot \left[ {3x + {9^0}} \right] = \dfrac{{\sqrt 3 }}{3}\]
c] \[\sin \left[ {3x + 1} \right] = \sin \left[ {x - 2} \right]\]
d] \[\cos 3x - \sin 2x = 0\]
e] \[\tan \left[ {2x + 1} \right] + \cot x = 0\]
g] \[\sin \left[ {x - {{120}^0}} \right] + \cos 2x = 0\]
Bài 2: Tìm điều kiện xác định của hàm số \[y = \sqrt {\dfrac{{1 + {{\cot }^2}x}}{{1 - \sin 3x}}} \]
Lời giải chi tiết
Bài 1:
a] \[\tan \left[ {2x - 1} \right] = \sqrt 3 \][1]
ĐK: \[\cos [2x - 1] \ne 0\]
\[\Leftrightarrow 2x - 1 \ne \dfrac{\pi }{2} + k\pi \]
\[\Leftrightarrow x \ne \dfrac{\pi }{4} + \dfrac{1}{2} + k\dfrac{\pi }{2}\]
Pt [1] \[\,\,\, \Leftrightarrow \tan [2x - 1] = \tan \dfrac{\pi }{3}\]
\[\Leftrightarrow 2x - 1 = \dfrac{\pi }{3} + k\pi \]
\[\Leftrightarrow x = \dfrac{\pi }{6} + \dfrac{1}{2} + k\dfrac{\pi }{2}\,[k \in \mathbb{Z}]\,[tmdk]\]
b] \[\cot \left[ {3x + {9^0}} \right] = \dfrac{{\sqrt 3 }}{3}\] [2]
ĐK: \[\sin [3x + {9^0}] \ne 0 \]
\[\Leftrightarrow 3x + {9^0} \ne k{180^0}\]
\[\Leftrightarrow x \ne - {3^0} + k{60^0}\]
[2] \[ \Leftrightarrow \cot [3x + {9^0}] = \cot {60^0}\]
\[\Leftrightarrow 3x + {9^0} = {60^0} + k{180^0}\]
\[\Leftrightarrow x = {17^0} + k{60^0}\,[k \in \mathbb{Z}]\]
c] \[\sin \left[ {3x + 1} \right] = \sin \left[ {x - 2} \right]\]
\[\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x + 1 = x - 2 + k2\pi }\\{3x + 1 = \pi - x + 2 + k2\pi }\end{array}} \right.\]
\[\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{{ - 3}}{2} + k\pi }\\{x = \dfrac{\pi }{4} + \dfrac{1}{4} + k\dfrac{\pi }{2}}\end{array}} \right.[k \in \mathbb{Z}]\]
d] \[\cos 3x - \sin 2x = 0 \]
\[\Leftrightarrow \cos 3x = \sin 2x \]
\[\Leftrightarrow \cos 3x = \cos [\dfrac{\pi }{2} - 2x]\]
\[ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \dfrac{\pi }{2} - 2x + k2\pi }\\{3x = 2x - \dfrac{\pi }{2} + k2\pi }\end{array}} \right. \]
\[\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{{10}} + k\dfrac{{2\pi }}{5}}\\{x = - \dfrac{\pi }{2} + k2\pi }\end{array}} \right.\,\,[k \in \mathbb{Z}]\]
e] \[\tan \left[ {2x + 1} \right] + \cot x = 0\] [3]
ĐK: \[\left\{ {\begin{array}{*{20}{c}}{\cos [2x + 1] \ne 0}\\{\sin x \ne 0}\end{array}} \right. \]
\[\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{2x + 1 \ne \dfrac{\pi }{2} + k\pi }\\{x \ne k\pi }\end{array}} \right. \]
\[\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x \ne \dfrac{\pi }{4} - \dfrac{1}{2} + k\dfrac{\pi }{2}}\\{x \ne k\pi }\end{array}} \right.\]
\[[3]\, \Leftrightarrow \tan [2x + 1] = - \cot x \]
\[\Leftrightarrow \tan [2x + 1] = \tan [\dfrac{\pi }{2} + x] \]
\[\Leftrightarrow 2x + 1 = \dfrac{\pi }{2} + x + k\pi \]
\[\Leftrightarrow x = \dfrac{\pi }{2} - 1 + k\pi \,[k \in \mathbb{Z}]\]
g] \[\sin \left[ {x - {{120}^0}} \right] + \cos 2x = 0\]
\[\Leftrightarrow \cos 2x = - \sin [x - {120^0}]\]
\[\Leftrightarrow \cos 2x = \cos [{90^0} + x - {120^0}]\]
\[ \Leftrightarrow \cos 2x = \cos \left[ { - {{30}^0} + x} \right]\]
\[ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x = - {{30}^0} + x + k{{360}^0}}\\{2x = {{30}^0} - x + k{{360}^0}}\end{array} } \right.\]
\[\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - {{30}^0} + k{{360}^0}}\\{x = {{10}^0} + k{{120}^0}}\end{array}[k \in \mathbb{Z}]} \right.\]
Bài 2: \[y = \sqrt {\dfrac{{1 + {{\cot }^2}x}}{{1 - \sin 3x}}} \]
ĐK: \[\left\{ {\begin{array}{*{20}{c}}{1 - \sin 3x \ne 0}\\{\sin x \ne 0}\\{\,\,\,\,\,\dfrac{{1 + {{\cot }^2}x}}{{1 - \sin 3x}} \ge 0\,\,\,\,}\end{array}} \right.\]
\[\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{1 - \sin 3x \ne 0}\\{\sin x \ne 0}\end{array}} \right. \]
\[ \Leftrightarrow \left\{ \begin{array}{l}
\sin 3x \ne 1\\
\sin x \ne 0
\end{array} \right.\]
\[\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{3x \ne \dfrac{\pi }{2} + k2\pi }\\{x \ne k\pi }\end{array}} \right. \]
\[\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x \ne \dfrac{\pi }{6} + k\dfrac{{2\pi }}{3}}\\{x \ne k\pi }\end{array}} \right.\]