Bài tập 46 trang 48 SGK Toán 11 NC

a)

\(\begin{array}{l}
\sin \left( {x - \frac{{2\pi }}{3}} \right) = \cos 2x\\
 \Leftrightarrow \sin \left( {x - \frac{{2\pi }}{3}} \right) = \sin \left( {\frac{\pi }{2} - 2x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \frac{{2\pi }}{3} = \frac{\pi }{2} - 2x + k2\pi \\
x - \frac{{2\pi }}{3} = \pi  - \frac{\pi }{2} + 2x + k2\pi \,
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{7\pi }}{8} + k\frac{{2\pi }}{3}\\
x =  - \frac{{7\pi }}{6} - k2\pi 
\end{array} \right.
\end{array}\)

b) Với ĐKXĐ của phương trình ta có tan(2x+450) = cot(450−2x) và 

\(\tan \left( {{{180}^0} - \frac{x}{2}} \right) = \tan \left( { - \frac{x}{2}} \right)\) nên:

\(\begin{array}{l}
\tan \left( {2x + {{45}^0}} \right)\tan \left( {{{180}^0} - \frac{x}{2}} \right) = 1\\
 \Leftrightarrow \cot \left( {{{45}^0} - 2x} \right)\tan \left( { - \frac{x}{2}} \right) = 1\\
 \Leftrightarrow \tan \left( { - \frac{x}{2}} \right) = \tan \left( {{{45}^0} - 2x} \right)\\
 \Leftrightarrow  - \frac{x}{2} = {45^0} - 2x + k{180^0}\\
 \Leftrightarrow x = {30^0} + k{120^0},k \in Z
\end{array}\)

c)

\(\begin{array}{l}
\cos 2x - {\sin ^2}x = 0\\
 \Leftrightarrow \cos 2x - \frac{{1 - \cos 2x}}{2} = 0\\
 \Leftrightarrow 3\cos 2x - 1 = 0\\
 \Leftrightarrow \cos 2x = \frac{1}{3}\\
 \Leftrightarrow \cos 2x = \cos \alpha \,\,\left( {\cos \alpha  = \frac{1}{3}} \right)\\
 \Leftrightarrow x =  \pm \frac{\alpha }{2} + k\pi \left( {k \in Z} \right)
\end{array}\)

d)

\(\begin{array}{*{20}{l}}
{5\tan x - 2\cot x = 3}\\
{ \Leftrightarrow 5\tan x - \frac{2}{{\tan x}} = 3}\\
{ \Leftrightarrow 3{{\tan }^2}x - 3\tan x - 2 = 0}\\
\begin{array}{l}
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\tan x = 1}\\
{\tan x =  - \frac{2}{5}}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = \frac{\pi }{4} + k\pi }\\
{x = \alpha  + k\pi }
\end{array}} \right.\left( {k \in Z} \right)
\end{array}
\end{array}\)

trong đó \(\tan \alpha  =  - \frac{2}{5}\).

-- Mod Toán 11 HỌC247