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 $y = \sqrt {\dfrac{{1 + {{\cot }^2}x}}{{1 - \sin 3x}}}$

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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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