Đề kiểm tra 15 phút – Chương 1 – Đề số 5 – Đại số và giải tích 11

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

\(\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{1 - \sin 3x \ne 0}\\{\sin x \ne 0}\end{array}} \right. \)

\( \Leftrightarrow \left\{ \begin{array}{l}
\sin 3x \ne 1\\
\sin x \ne 0
\end{array} \right.\)

\(\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{3x \ne \dfrac{\pi }{2} + k2\pi }\\{x \ne k\pi }\end{array}} \right. \)

\(\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x \ne \dfrac{\pi }{6} + k\dfrac{{2\pi }}{3}}\\{x \ne k\pi }\end{array}} \right.\)

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