Bài tập 6 trang 224 SGK Toán 11 NC

Bài tập 6 trang 224 SGK Toán 11 NC

a. Đặt \(t = \frac{1}{{\cos x}}\left( {x \ne \frac{\pi }{2} + k\pi } \right)\)

Ta có:

\(\begin{array}{l}
2({t^2} - 1) + 3 = 3t\\
 \Leftrightarrow 2{t^2} - 3t + 1 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
t = 1\\
t = \frac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = 2\left( l \right)
\end{array} \right.\\
 \Leftrightarrow x = k2\pi 
\end{array}\)

b. Điều kiện : \(\cos x \ne 0 \Leftrightarrow x = \frac{\pi }{2} + k\pi \)

\(\begin{array}{l}
{\tan ^2}x = \frac{{1 + \cos x}}{{1 + \sin x}}\\
 \Leftrightarrow \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} = \frac{{1 + \cos x}}{{1 + \sin x}}\\
 \Leftrightarrow \frac{{1 - {{\cos }^2}x}}{{1 - {{\sin }^2}x}} = \frac{{1 + \cos x}}{{1 + \sin x}}\\
 \Leftrightarrow \frac{{1 - {{\cos }^2}x}}{{1 - \sin x}} = 1 + \cos x\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x =  - 1\\
1 - \cos x = 1 - \sin x
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x =  - 1\\
\tan x = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \pi  + k2\pi \\
x = \frac{\pi }{4} + k\pi 
\end{array} \right.(k \in Z)
\end{array}\)

c. Điều kiện 

\(\begin{array}{l}
\cos x \ne 0,\cos 2x \ne 0\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{\cos x \ne 0}\\
{cosx \ne  \pm \frac{1}{{\sqrt 2 }}}
\end{array}} \right.
\end{array}\)

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

-- Mod Toán 11