Bài 7 trang 232 SBT đại số và giải tích 11

Bài làm:

Giải các phương trình

LG a

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

Lời giải chi tiết:

\(\begin{array}{l}{\cos ^2}x + {\cos ^2}2x - {\cos ^2}3x - {\cos ^2}4x = 0\\ \Leftrightarrow \frac{{1 + \cos 2x}}{2} + \frac{{1 + \cos 4x}}{2} - \frac{{1 + \cos 6x}}{2} - \frac{{1 + \cos 8x}}{2} = 0\\ \Leftrightarrow 1 + \cos 2x + 1 + \cos 4x - 1 - \cos 6x - 1 - \cos 8x = 0\\ \Leftrightarrow \left( {\cos 2x + \cos 4x} \right) - \left( {\cos 6x + \cos 8x} \right) = 0\\ \Leftrightarrow 2\cos 3x\cos x - 2\cos 7x\cos x = 0\\ \Leftrightarrow 2\cos x\left( {\cos 3x - \cos 7x} \right) = 0\\ \Leftrightarrow 2\cos x.\left[ { - 2\sin 5x\sin \left( { - 2x} \right)} \right] = 0\\ \Leftrightarrow 4\cos x\sin 5x\sin 2x = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 0\\\sin 5x = 0\\\sin 2x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\5x = k\pi \\2x = k\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\x = \frac{{k\pi }}{5}\\x = \frac{{k\pi }}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{k\pi }}{5}\\x = \frac{{k\pi }}{2}\end{array} \right.,k \in \mathbb{Z}\end{array}\)

LG b

\(\cos 4x\cos \left( {\pi  + 2x} \right) - \sin 2x\cos \left( {\frac{\pi }{2} - 4x} \right) \) \(= \frac{{\sqrt 2 }}{2}\sin 4x\)

Lời giải chi tiết:

\(\begin{array}{l}\cos 4x\cos \left( {\pi  + 2x} \right) - \sin 2x\cos \left( {\frac{\pi }{2} - 4x} \right) = \frac{{\sqrt 2 }}{2}\sin 4x\\ \Leftrightarrow  - \cos 4x\cos 2x - \sin 2x\sin 4x = \frac{{\sqrt 2 }}{2}\sin 4x\\ \Leftrightarrow  - \left( {\cos 4x\cos 2x + \sin 2x\sin 4x} \right) = \frac{{\sqrt 2 }}{2}\sin 4x\\ \Leftrightarrow  - \cos \left( {4x - 2x} \right) = \frac{{\sqrt 2 }}{2}.2\sin 2x\cos 2x\\ \Leftrightarrow  - \cos 2x = \sqrt 2 \sin 2x\cos 2x\\ \Leftrightarrow \sqrt 2 \sin 2x\cos 2x + \cos 2x = 0\\ \Leftrightarrow \cos 2x\left( {\sqrt 2 \sin 2x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = 0\\\sqrt 2 \sin 2x + 1 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{2} + k\pi \\\sin 2x =  - \frac{1}{{\sqrt 2 }}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\2x =  - \frac{\pi }{4} + k2\pi \\2x = \frac{{5\pi }}{4} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\x =  - \frac{\pi }{8} + k\pi \\x = \frac{{5\pi }}{8} + k\pi \end{array} \right.,k \in \mathbb{Z}\end{array}\)

LG c

\(\tan \left( {{{120}^0} + 3x} \right) - \tan \left( {{{140}^0} - x} \right) = 2\sin \left( {{{80}^0} + 2x} \right)\)

Lời giải chi tiết:

ĐK:

\(\begin{array}{l}\left\{ \begin{array}{l}{120^0} + 3x \ne {90^0} + k{.180^0}\\{140^0} - x \ne {90^0} + k{.180^0}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3x \ne  - {30^0} + k{.180^0}\\x \ne {50^0} - k{.180^0}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ne  - {10^0} + k{.180^0}\\x \ne {50^0} - k{.180^0}\end{array} \right.\end{array}\)

\(\begin{array}{l}\tan \left( {{{120}^0} + 3x} \right) - \tan \left( {{{140}^0} - x} \right) = 2\sin \left( {{{80}^0} + 2x} \right)\\ \Leftrightarrow \tan \left[ {3\left( {{{40}^0} + x} \right)} \right] - \tan \left[ {{{180}^0} - \left( {{{40}^0} + x} \right)} \right] = 2\sin \left[ {2\left( {{{40}^0} + x} \right)} \right]\\ \Leftrightarrow \tan \left[ {3\left( {{{40}^0} + x} \right)} \right] - \tan \left[ { - \left( {{{40}^0} + x} \right)} \right] = 2\sin \left[ {2\left( {{{40}^0} + x} \right)} \right]\\ \Leftrightarrow \tan \left[ {3\left( {{{40}^0} + x} \right)} \right] + \tan \left( {{{40}^0} + x} \right) = 2\sin \left[ {2\left( {{{40}^0} + x} \right)} \right]\end{array}\)

Đặt \({40^0} + x = y\) ta được:

\(\begin{array}{l}\tan 3y + \tan y = 2\sin 2y\\ \Leftrightarrow \frac{{\sin 3y}}{{\cos 3y}} + \frac{{\sin y}}{{\cos y}} = 2\sin 2y\\ \Leftrightarrow \frac{{\sin 3y\cos y + \sin y\cos 3y}}{{\cos 3y\cos y}} = \frac{{2\sin 2y\cos 3y\cos y}}{{\cos 3y\cos y}}\\ \Rightarrow \sin 3y\cos y + \sin y\cos 3y = 2\sin 2y\cos 3y\cos y\\ \Leftrightarrow \sin 4y - 2\sin 2y\cos 3y\cos y = 0\\ \Leftrightarrow 2\sin 2y\cos 2y - 2\sin 2y\cos 3y\cos y = 0\\ \Leftrightarrow 2\sin 2y\left( {\cos 2y - \cos 3y\cos y} \right) = 0\\ \Leftrightarrow 2\sin 2y\left[ {\cos 2y - \frac{1}{2}\left( {\cos 4y + \cos 2y} \right)} \right] = 0\\ \Leftrightarrow 2\sin 2y\left( {\frac{1}{2}\cos 2y - \frac{1}{2}\cos 4y} \right) = 0\\ \Leftrightarrow \sin 2y\left( {\cos 2y - \cos 4y} \right) = 0\\ \Leftrightarrow \sin 2y.\left[ { - 2\sin 3y\sin \left( { - y} \right)} \right] = 0\\ \Leftrightarrow 2\sin y\sin 2y\sin 3y = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin y = 0\\\sin 2y = 0\\\sin 3y = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}y = k{180^0}\\2y = k{180^0}\\3y = k{180^0}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}y = k{.180^0}\\y = k{.90^0}\\y = k{.60^0}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}y = k{.60^0}\\y = {90^0} + k{.180^0}\end{array} \right.\end{array}\)

Suy ra

\(\begin{array}{l}\left[ \begin{array}{l}x + {40^0} = k{.60^0}\\x + {40^0} = {90^0} + k{.180^0}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  - {40^0} + k{.60^0}\\x = {50^0} + k{180^0}\left( {loai} \right)\end{array} \right.\\ \Leftrightarrow x =  - {40^0} + k{.60^0},k \in \mathbb{Z}\end{array}\)

Vậy pt có nghiệm \(x =  - {40^0} + k{.60^0},k \in \mathbb{Z}\).

LG d

\({\tan ^2}\frac{x}{2} + {\sin ^2}\frac{x}{2}\tan \frac{x}{2} + {\cos ^2}\frac{x}{2}.{\cot ^2}\frac{x}{2} + {\cot ^2}\frac{x}{2} + \sin x = 4\)

Lời giải chi tiết:

x = (4k + 1) π/2;

x = (-1)^(k+1)arcsin2/3 + kπ.

LG e

\(\frac{{\sin 2t + 2{{\cos }^2}t - 1}}{{\cos t - \cos 3t + \sin 3t - \sin t}} = \cos t\)

Lời giải chi tiết:

\(\begin{array}{l}\frac{{\sin 2t + 2{{\cos }^2}t - 1}}{{\cos t - \cos 3t + \sin 3t - \sin t}} = \cos t\\ \Leftrightarrow \frac{{\sin 2t + \cos 2t}}{{ - 2\sin 2t\sin \left( { - t} \right) + 2\cos 2t\sin t}} = \cos t\\ \Leftrightarrow \frac{{\sin 2t + \cos 2t}}{{2\sin 2t\sin t + 2\cos 2t\sin t}} = \cos t\\ \Leftrightarrow \frac{{\sin 2t + \cos 2t}}{{2\sin t\left( {\sin 2t + \cos 2t} \right)}} = \cos t\end{array}\)

ĐK: \(\left\{ \begin{array}{l}\sin t \ne 0\\\sin 2t + \cos 2t \ne 0\end{array} \right.\)(*)

\(\begin{array}{l} \Leftrightarrow \frac{1}{{2\sin t}} = \cos t\\ \Rightarrow 1 = 2\sin t\cos t\\ \Leftrightarrow 1 = \sin 2t\\ \Leftrightarrow 2t = \frac{\pi }{2} + k2\pi \\ \Leftrightarrow t = \frac{\pi }{4} + k\pi \left( {TM\,\,\,\left( * \right)} \right)\end{array}\)

Vậy pt có nghiệm \(t = \frac{\pi }{4} + k\pi ,k \in \mathbb{Z}\).

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