Chọn A

Ta có:

\(\begin{array}{l}cos2x = cos\left( {x + \frac{\pi }{3}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = x + \frac{\pi }{3} + k2\pi \\2x =  - x - \frac{\pi }{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x =  - \frac{\pi }{9} + k\frac{{2\pi }}{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

Với \(x = \frac{\pi }{3} + k2\pi \),\(k \in \mathbb{Z}\) đạt giá trị âm lớn nhất khi k = – 1, khi đó \(x = \frac{\pi }{3} - 2\pi  = \frac{{ - 5\pi }}{3}\)

Với \(x =  - \frac{\pi }{9} + k\frac{{2\pi }}{3}\),\(k \in \mathbb{Z}\) đạt giá trị âm lớn nhất khi k = 0, khi đó \(x = x =  - \frac{\pi }{9} + 0.\frac{{2\pi }}{3} =  - \frac{\pi }{9}\)

Vậy nghiệm âm lớn nhất của phương trình đã cho là \( - \frac{\pi }{9}\).
Đáp án: A

a) Vì \(\sin \frac{\pi }{6} = \frac{1}{2}\) nên ta có phương trình \(sin2x = \sin \frac{\pi }{6}\)

\( \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{6} + k2\pi \\2x = \pi  - \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \\x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

\(\begin{array}{l}b,\,\,sin(x - \frac{\pi }{7}) = sin\frac{{2\pi }}{7}\\ \Leftrightarrow \left[ \begin{array}{l}x - \frac{\pi }{7} = \frac{{2\pi }}{7} + k2\pi \\x - \frac{\pi }{7} = \pi  - \frac{{2\pi }}{7} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{3\pi }}{7} + k2\pi \\x = \frac{{6\pi }}{7} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

\(\begin{array}{l}\;c)\;sin4x - cos\left( {x + \frac{\pi }{6}} \right) = 0\\ \Leftrightarrow sin4x = cos\left( {x + \frac{\pi }{6}} \right)\\ \Leftrightarrow sin4x = \sin \left( {\frac{\pi }{2} - x - \frac{\pi }{6}} \right)\\ \Leftrightarrow sin4x = \sin \left( {\frac{\pi }{3} - x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}4x = \frac{\pi }{3} - x + k2\pi \\4x = \pi  - \frac{\pi }{3} + x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{15}} + k\frac{{2\pi }}{5}\\x = \frac{{2\pi }}{9} + k\frac{{2\pi }}{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

Lời giải:

$\tan (\frac{\pi}{2}+x)-3\tan ^2x=\frac{\cos 2x-1}{\cos ^2x}=\frac{2\cos ^2x-2}{\cos ^2x}=\frac{2(\cos ^2x-1)}{\cos ^2x}$

$=\frac{-2\sin ^2x}{\cos ^2x}=-2\tan ^2x$

$\Leftrightarrow \tan (x+\frac{\pi}{2})=\tan ^2x$

Dễ thấy $\tan x=0$ không thỏa mãn nên $\tan x\neq 0$. Do đó pt $\Leftrightarrow \tan ^2x=\tan [\pi +(x-\frac{\pi}{2})]=\tan (x-\frac{\pi}{2})=-\tan (\frac{\pi}{2}-x)=-\cot x =\frac{-1}{\tan x}$

$\Rightarrow \tan ^3x=-1$

$\Rightarrow \tan x=-1$

$\Rightarrow x=\frac{-\pi}{4}+k\pi$ với $k$ nguyên.

ĐKXĐ: \(x\ne\frac{\pi}{6}+\frac{k\pi}{3}\)

\(\Leftrightarrow\frac{cos^2x-cos3x.cos5x}{cos3x.cosx}-4\left[1-2sin^2\left(2x+\frac{11\pi}{2}\right)\right]-4cos2x=0\)

\(\Leftrightarrow\frac{2cos^2x-cos2x-cos8x}{cos4x+cos2x}-4cos\left(4x+11\pi\right)-4cos2x=0\)

\(\Leftrightarrow\frac{1-cos8x}{cos4x+cos2x}+4cos4x-4cos2x=0\)

\(\Leftrightarrow1-cos8x+4\left(cos4x-cos2x\right)\left(cos4x+cos2x\right)=0\)

\(\Leftrightarrow1-cos8x+4cos^24x-4cos^22x=0\)

\(\Leftrightarrow1-\left(2cos^24x-1\right)+4cos^24x-2\left(1+cos4x\right)=0\)

\(\Leftrightarrow cos^24x-cos4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x=1\end{matrix}\right.\) \(\Leftrightarrow...\)

@Nguyễn Việt Lâm giúp em với ạ

a.

\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)

\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

b.

\(sinx\left(1+2cos^2x-1\right)+2sinx.cosx=1+2cos^2x-1\)

\(\Leftrightarrow cos^2x.sinx+sinx.cosx-cos^2x=0\)

\(\Leftrightarrow cosx\left(sinx.cosx+sinx-cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\sinx.cosx+sinx-cosx=0\left(1\right)\end{matrix}\right.\)

Xét (1), đặt \(sinx-cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)

\(\Rightarrow\frac{1-t^2}{2}+t=0\)

\(\Leftrightarrow-t^2+2t+1=0\Rightarrow\left[{}\begin{matrix}t=1-\sqrt{2}\\t=1+\sqrt{2}>\sqrt{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1-\sqrt{2}\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

mai đăng lại bài này nhé t làm cho h đi ngủ

a) \(\cos \left( {3x - \frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \cos \left( {3x - \frac{\pi }{4}} \right) = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi }\\{3x - \frac{\pi }{4} =  - \frac{{3\pi }}{4} + k2\pi }\end{array}} \right.\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \pi  + k2\pi }\\{3x =  - \frac{\pi }{2} + k2\pi }\end{array}} \right.\)

\( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\\{x =  - \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)

b) \(2{\sin ^2}x - 1 + \cos 3x = 0\;\;\;\;\; \Leftrightarrow \cos 2x + \cos 3x = 0\;\; \Leftrightarrow 2\cos \frac{{5x}}{2}\cos \frac{x}{2} = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos \frac{{5x}}{2} = 0}\\{\cos \frac{x}{2} = 0}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\frac{{5x}}{2} = \frac{\pi }{2} + k\pi }\\{\frac{{5x}}{2} =  - \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = \frac{\pi }{2} + k\pi }\\{\frac{x}{2} =  - \frac{\pi }{2} + k\pi }\end{array}} \right.\;\;\;\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x =  - \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = \pi  + k2\pi }\\{x =  - \pi  + k2\pi }\end{array}} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)

c) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x =  - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)\)