Đề kiểm tra 15 phút – Chương 1 – Đề số 8 – Đại số và giải tích 11

Đề bài

Bài 1: Giải các phương trình sau:

a) \(\cos 2x + 9\cos x + 5 = 0\)

b) \({\left( {\sin x - \cos x} \right)^2} \)\(- \left( {\sqrt 2  + 1} \right)(\sin x - \cos x) + \sqrt 2  = 0\)

c) \(2{\sin ^2}x + \sqrt 3 \sin 2x = 3\)

d) \( - \sin x - \cos x = \sqrt 2 \sin 5x\)

Bài 2: Giải các phương trình sau:

a) \(2{\sin ^2}x + \left( {1 - \sqrt 3 } \right)\sin x\cos x \)\(+ \left( {1 - \sqrt 3 } \right){\cos ^2}x = 1\)

b) \(1 + 2\sin x\cos x = \sin x + 2\cos x\)

Lời giải chi tiết

Bài 1:

\(a)\,\cos 2x + 9\cos x + 5 = 0 \)

\(\Leftrightarrow 2{\cos ^2}x - 1 + 9\cos x + 5 = 0\)

\(\Leftrightarrow 2{\cos ^2}x + 9\cos  + 4 = 0  (1)\)

Đặt: \(\cos x = t  ( - 1 \le t \le 1)\)

Khi đó (1) trở thành: \(2{t^2} + 9t + 4 = 0 \)

\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{t = \dfrac{{ - 1}}{2}\,(tm)}\\{t =  - 4\,\,(ktm)}\end{array}} \right.\)

Với \(t = \dfrac{{ - 1}}{2} \Rightarrow \cos x = \dfrac{{ - 1}}{2}\)

\(\Leftrightarrow \cos x = \cos \dfrac{{2\pi }}{3} \Leftrightarrow x =  \pm \dfrac{{2\pi }}{3} + k2\pi \)

\(b)\,{\left( {\sin x - \cos x} \right)^2}\)\( - \left( {\sqrt 2  + 1} \right)(\sin x - \cos x) + \sqrt 2  = 0  (2)\)

Đặt: \(\sin x - \cos x = t  ( - \sqrt 2  \le t \le \sqrt 2 )\)

Khi đó (2) trở thành: \({t^2} - (\sqrt 2  + 1)t + \sqrt 2  = 0 \)

\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{t = \sqrt 2 \,\,\,\,(tm)}\\{t = 1\,\,\,(tm)}\end{array}} \right.\)

Với \(t = \sqrt 2 \)

\(\begin{array}{l} \Rightarrow \sin x - \cos x = \sqrt 2 \\ \Leftrightarrow \sqrt 2 \sin (x - \dfrac{\pi }{4}) = \sqrt 2 \\ \Leftrightarrow \sin (x - \dfrac{\pi }{4}) = 1\\ \Leftrightarrow x - \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\ \Leftrightarrow x = \dfrac{{3\pi }}{4} + k2\pi \end{array}\)

Với \(t = 1\)

\(\begin{array}{l} \Rightarrow \sin x - \cos x = 1\\ \Leftrightarrow \sin (x - \dfrac{\pi }{4}) = \dfrac{1}{{\sqrt 2 }}\\ \Leftrightarrow \sin (x - \dfrac{\pi }{4}) = \sin \dfrac{\pi }{4}\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x - \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi }\\{x - \dfrac{\pi }{4} = \pi  - \dfrac{\pi }{4} + k2\pi }\end{array}\\ } \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{2} + k2\pi }\\{x = \pi  + k2\pi }\end{array}} \right.\end{array}\)

\(\begin{array}{l}c)\,\,\,\,\,\,2{\sin ^2}x + \sqrt 3 \sin 2x = 3\\ \Leftrightarrow 2\dfrac{{1 - 2\cos x}}{2} + \sqrt 3 \sin 2x = 3\\ \Leftrightarrow 1 - \cos 2x + \sqrt 3 \sin 2x = 3\\ \Leftrightarrow \sqrt 3 \sin 2x - \cos 2x = 2\\ \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin 2x - \dfrac{1}{2}\cos 2x = 1\\ \Leftrightarrow \cos \dfrac{\pi }{6}\sin 2x - \sin \dfrac{\pi }{6}\cos 2x = 1\\ \Leftrightarrow \sin (2x - \dfrac{\pi }{6}) = 1\\ \Leftrightarrow 2x - \dfrac{\pi }{6} = \dfrac{\pi }{2} = k2\pi \\ \Leftrightarrow x = \dfrac{\pi }{3} + k\pi \end{array}\)

\(\begin{array}{l}d)\,\,\,\,\,\,\, - \sin x - \cos x = \sqrt 2 \sin 5x \\ \Leftrightarrow \dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x =  - \sin 5x\\ \Leftrightarrow \cos \dfrac{\pi }{4}\sin x + \sin \dfrac{\pi }{4}\cos x = \sin ( - 5x)\\ \Leftrightarrow \sin (x + \dfrac{\pi }{4}) = \sin ( - 5x)\\  \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x + \dfrac{\pi }{4} =  - 5x + k2\pi }\\{x + \dfrac{\pi }{4} = \pi  + 5x + k2\pi }\end{array}  } \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{6x =  - \dfrac{\pi }{4} + k2\pi }\\{ - 4x = \dfrac{{3\pi }}{4} + k2\pi }\end{array}  } \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x =  - \dfrac{\pi }{{24}} + k\dfrac{\pi }{3}}\\{x =  - \dfrac{{3\pi }}{{16}} - k\dfrac{\pi }{2}}\end{array}} \right.\,\,\,(k \in \mathbb{Z})\end{array}\)

Bài 2:

\(a)\,2{\sin ^2}x + \left( {1 - \sqrt 3 } \right)\sin x\cos x \)\(+ \left( {1 - \sqrt 3 } \right){\cos ^2}x = 1  (1)\)

Với \(\cos x = 0 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \,\,(k \in \mathbb{Z})\)

Thay \(x = \dfrac{\pi }{2} + k\pi \,\,\) vào phương trình (1) ta được:  2=1 ( vô lí) \( \Rightarrow x = \dfrac{\pi }{2} + k\pi \,\,(k \in \mathbb{Z})\) không là nghiệm của (1).

Với \(\cos x \ne 0 \)

\(\Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \,\,(k \in \mathbb{Z})\)

\(\begin{array}{l}(1) \Leftrightarrow 2\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + (1 - \sqrt 3 )\dfrac{{\sin x}}{{\cos x}} + (1 - \sqrt 3 ) = \dfrac{1}{{{{\cos }^2}x}}\\ \Leftrightarrow 2{\tan ^2}x - (1 - \sqrt 3 )\tan x + (1 - \sqrt 3 ) = 1 + {\tan ^2}x\\ \Leftrightarrow {\tan ^2}x + (1 - \sqrt 3 )\tan x - \sqrt 3  = 0 \\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\tan x = \sqrt 3 }\\{\tan x =  - 1}\end{array}} \right. \\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\tan x = \tan \dfrac{\pi }{3}}\\{\tan x = \tan ( - \dfrac{\pi }{4})}\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{3} + k\pi }\\{x = \dfrac{{ - \pi }}{4} + k\pi }\end{array}} \right.\,\,(k \in \mathbb{Z})\end{array}\)

Vậy phương trình có nghiệm là: \(\,x = \dfrac{\pi }{3} + k\pi ;x = \dfrac{{ - \pi }}{4} + k\pi (k \in \mathbb{Z})\)

\(\begin{array}{l}b)\,\,\,\,\,1 + 2\sin x\cos x = \sin x + 2\cos x \\ \Leftrightarrow 1 + 2\sin x\cos x - \sin x - 2\cos x = 0\\ \Leftrightarrow (2\sin x\cos x - \sin x) - (2\cos x - 1) = 0\\ \Leftrightarrow \sin x(2\cos x - 1) - (2\cos x - 1) = 0\\ \Leftrightarrow (2\cos x - 1)(\sin x - 1) = 0\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2\cos x - 1 = 0}\\{\sin x - 1 = 0}\end{array}\left[ {\begin{array}{*{20}{c}}{\cos x = \dfrac{1}{2}}\\{\sin x = 1}\end{array}} \right.} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos x = \cos \dfrac{\pi }{6}}\\{\sin x = \sin \dfrac{\pi }{2}}\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x =\pm \dfrac{\pi }{6} + k2\pi }\\{x = \dfrac{\pi }{2} + k2\pi }\end{array}\,\,(k \in \mathbb{Z})} \right.\end{array}\)

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