Đề kiểm tra 15 phút – Chương 1 – Đề số 9 – Đại số và giải tích 11

Đề bài

Bài 1: Giải các phương trình sau:

a) \(2\cos (2x - \dfrac{\pi }{5}) = 1\)

b) \(\sin \left( {x - \dfrac{\pi }{3}} \right) = \sin \left( {2x + \dfrac{\pi }{6}} \right)\)

c) \(\sin 3x + \sin 5x = 0\)

d) \(3\tan 4x - 2\cot 4x + 1 = 0\)

Bài 2: Tìm \(x \in {\rm{[}}0;14]\) nghiệm đúng phương trình:

\(\cos 3x - 4\cos 2x + 3\cos x - 4 = 0\)

Lời giải chi tiết

Bài 1:

\(\begin{array}{l}a)  2\cos (2x - \dfrac{\pi }{5}) = 1\\ \Leftrightarrow \cos (2x - \dfrac{\pi }{5}) = \dfrac{1}{2}\\ \Leftrightarrow \cos (2x - \dfrac{\pi }{5}) = \cos \dfrac{\pi }{3}\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x - \dfrac{\pi }{5} = \dfrac{\pi }{3} + k2\pi }\\{2x - \dfrac{\pi }{5} = \dfrac{{ - \pi }}{3} + k2\pi }\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{{4\pi }}{{15}} + k\pi }\\{x = \dfrac{{ - \pi }}{{15}} + k\pi }\end{array}} \right.\end{array}\)

\(b) \; \sin \left( {x - \dfrac{\pi }{3}} \right) = \sin \left( {2x + \dfrac{\pi }{6}} \right) \\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x + \dfrac{\pi }{6} = x - \dfrac{\pi }{3} + k2\pi }\\{2x + \dfrac{\pi }{6} = \pi  - x + \dfrac{\pi }{3} + k2\pi }\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \dfrac{{ - \pi }}{2} + k2\pi }\\{x = \dfrac{{7\pi }}{{18}} + k\dfrac{{2\pi }}{3}}\end{array}} \right.\)

\(\begin{array}{l}c)\;  \sin 3x + \sin 5x = 0 \\\Leftrightarrow \sin 5x =  - \sin 3x \\ \Leftrightarrow \sin 5x = \sin ( - 3x)\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{5x =  - 3x + k2\pi }\\{5x = \pi  + 3x + k2\pi }\end{array}} \right.\\ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = k\dfrac{\pi }{4}}\\{x = \dfrac{\pi }{2} + k\pi }\end{array}} \right.\end{array}\)

\(d)\;  3\tan 4x - 2\cot 4x + 1 = 0\,\,\,\,(1)\)

ĐK: \(\left\{ {\begin{array}{*{20}{c}}{\sin 4x \ne 0}\\{\cos 4x \ne 0}\end{array}} \right. \)

\(\Leftrightarrow \sin 8x \ne 0 \Leftrightarrow x \ne k\dfrac{\pi }{8}\)

Đặt \(\tan 4x = t(t \ne 0) \Rightarrow \cot 4x = \dfrac{1}{t}\)

Khi đó (1) trở thành: \(3t - \dfrac{2}{t} + 1 = 0\)

\(\Leftrightarrow 3{t^2} + t - 2 = 0 \)

\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{t =  - 1\,\,(TM)}\\{t = \dfrac{2}{3}\,(TM)}\end{array}} \right.\)

Với \(t =  - 1 \Rightarrow \tan 4x =  - 1\)

\(\Leftrightarrow \tan 4x = \tan \left( {\dfrac{{ - \pi }}{4}} \right)\)

\(\Leftrightarrow 4x = \dfrac{{ - \pi }}{4} + k\pi \)

\(\Leftrightarrow x = \dfrac{{ - \pi }}{{16}} + k\dfrac{\pi }{4}\,(TM)\)

Với \(t = \dfrac{2}{3} \Rightarrow \tan 4x = \dfrac{2}{3} \)

\(\Leftrightarrow x = \dfrac{1}{4}\arctan \dfrac{2}{3} + k\dfrac{\pi }{4}\,(TM)\)

Bài 2:

\(\begin{array}{l}\cos 3x - 4\cos 2x + 3\cos x - 4 = 0\\ \Leftrightarrow 4{\cos ^3}x - 3\cos x - 4(2{\cos ^2}x - 1) + 3\cos x - 4 = 0\\ \Leftrightarrow 4{\cos ^3}x - 3\cos x - 8{\cos ^2}x + 4 + 3\cos x - 4 = 0 \\\Leftrightarrow 4{\cos ^3}x - 8{\cos ^2}x = 0\\\Leftrightarrow {\cos ^3}x - 2\cos {}^2x = 0 \;\; (1)\end{array}\)

Đặt \(\,\cos x = t\,(\left| t \right| \le 1)\)

Khi đó (1) trở thành \({t^3} - 2{t^2} = 0\)

\(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{t = 0\,(TM)}\\{t = 2\,(KTM)}\end{array}} \right.\)

Với \(t = 0 \Rightarrow \cos x = 0 \)

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

Mà \(x \in {\rm{[}}0;14] \Rightarrow 0 \le \dfrac{\pi }{2} + k\pi  \le 14 \)\(\,\Leftrightarrow \dfrac{{ - 1}}{2} \le k \le \dfrac{{14}}{\pi } - \dfrac{1}{2}\)

Do \(k \in \mathbb{Z} \Rightarrow k \in \left\{ {0;1;2;3} \right\}\)

Vậy \(x = \dfrac{\pi }{2};x = \dfrac{{3\pi }}{2};x = \dfrac{{5\pi }}{2};x = \dfrac{{7\pi }}{2}\)