Bài tập1.52 trang 40 SBT Toán 11

ĐKXĐ: sinx ≠ 0; cosx ≠ 0 và tanx ≠ −1.

Ta có: \(\cot x = \frac{1}{{\tan x}}\);

$\begin{array}{l}\end{array}$\(\cos 2x = 2{\cos ^2}x - 1 = 2.\frac{1}{{{{\tan }^2}x + 1}} - 1 = \frac{{1 - {{\tan }^2}x}}{{{{\tan }^2}x + 1}}\)

\({{{\sin }^2}x = 1 - {{\cos }^2}x = 1 - \frac{1}{{{{\tan }^2}x + 1}} = \frac{{{{\tan }^2}x}}{{{{\tan }^2}x + 1}}}\)

\({ - \frac{1}{2}\sin 2x =  - \sin x\cos x =  - \frac{{\sin x}}{{\cos x}}{{\cos }^2}x =  - \tan x\frac{1}{{{{\tan }^2}x + 1}}}\)

Phương trình 

\(\begin{array}{l}
\cot x - 1 = \frac{{\cos 2x}}{{1 + \tan x}} + {\sin ^2}x - \frac{1}{2}\sin 2x\\
 \Leftrightarrow \frac{1}{{\tan x}} - 1 = \frac{{\frac{{1 - {{\tan }^2}x}}{{{{\tan }^2}x + 1}}}}{{1 + \tan x}} + \frac{{{{\tan }^2}x}}{{{{\tan }^2}x + 1}} - \frac{{\tan {\rm{x}}}}{{{{\tan }^2}x + 1}}
\end{array}\)

Đặt t = tanx ta được 

\(\begin{array}{l}
\frac{1}{t} - 1 = \frac{{\frac{{1 - {{\rm{t}}^2}}}{{{{\rm{t}}^2} + 1}}}}{{1 + {\rm{t}}}} + \frac{{{{\rm{t}}^2}}}{{{{\rm{t}}^2} + 1}} - \frac{{\rm{t}}}{{{{\rm{t}}^2} + 1}}\\
 \Leftrightarrow \frac{1}{t} - 1 = \frac{{1 - t}}{{{t^2} + 1}} + \frac{{{t^2} - t}}{{{t^2} + 1}}\\
 \Leftrightarrow \frac{{1 - t}}{t} = \frac{{1 - t}}{{{t^2} + 1}} + \frac{{t(t - 1)}}{{{t^2} + 1}}\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{1 - t = 0}\\
{\frac{1}{t} = \frac{1}{{{t^2} + 1}} - \frac{t}{{{t^2} + 1}}}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{t = 1}\\
{{t^2} + 1 = (1 - t)t}
\end{array}} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
t = 1\\
2{t^2} - t + 1 = 0\,\,\left( {VN} \right)
\end{array} \right.
\end{array}\)

\(t = 1 \Leftrightarrow \tan x = 1 \Leftrightarrow x = \frac{\pi }{4} + k\pi  \in Z\) $\begin{array}{l}\text{(thỏa mãn)}\end{array}$

Vậy phương trình có nghiệm là \(x = \frac{\pi }{4} + k\pi  \in Z\)

-- Mod Toán 11 HỌC247