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: \(\left\{ {\begin{array}{*{20}{c}}{1 - \sin 3x \ne 0}\\{\sin x \ne 0}\\{\,\,\,\,\,\dfrac{{1 + {{\cot }^2}x}}{{1 - \sin 3x}} \ge 0\,\,\,\,}\end{array}} \right.\)
\(\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{1 - \sin 3x \ne 0}\\{\sin x \ne 0}\end{array}} \right. \)
\( \Leftrightarrow \left\{ \begin{array}{l}
\sin 3x \ne 1\\
\sin x \ne 0
\end{array} \right.\)
\(\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{3x \ne \dfrac{\pi }{2} + k2\pi }\\{x \ne k\pi }\end{array}} \right. \)
\(\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x \ne \dfrac{\pi }{6} + k\dfrac{{2\pi }}{3}}\\{x \ne k\pi }\end{array}} \right.\)
