Đáp án:
Giải thích các bước giải:
13/
$\begin{array}{l}
1 + \tan x = 2\sqrt 2 \sin x\left( {ĐKXĐ:\cos x \ne 0 \Rightarrow x \ne \dfrac{\pi }{2} + k\pi } \right)\\
\Rightarrow 1 + \dfrac{{\sin x}}{{\cos x}} = 2\sqrt 2 \sin x\\
\Rightarrow \cos x + \sin x = 2\sqrt 2 \sin x.\cos x\\
\Rightarrow \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 .2\sin x.\cos x\\
\Rightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin 2x\\
\Rightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = 2x + k2\pi \\
x + \dfrac{\pi }{4} = \pi - 2x + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} - k2\pi \\
x = \dfrac{\pi }{4} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {tmdk} \right)
\end{array}$
14/
ĐKXĐ: $x \ne \dfrac{k\pi}{2}$
Ta có:
$\begin{array}{l}
\cot x - \tan x = \sin x + \cos x\\
\Leftrightarrow \dfrac{{\cos x}}{{\sin x}} - \dfrac{{\sin x}}{{\cos x}} = \sin x + \cos x\\
\Leftrightarrow \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x.\cos x}} = \sin x + \cos x\\
\Leftrightarrow \left( {\sin x + \cos x} \right)\left( {1 + \dfrac{{\sin x - \cos x}}{{\sin x.\cos x}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x + \cos x = 0\\
\dfrac{{\sin x - \cos x}}{{\sin x.\cos x}} = - 1
\end{array} \right.
\end{array}$
+) TH1:
$\begin{array}{l}
\sin x + \cos x = 0\\
\Leftrightarrow \cos \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\Leftrightarrow x - \dfrac{\pi }{4} = \dfrac{\pi }{2} + k\pi \\
\Leftrightarrow x = \dfrac{{3\pi }}{4} + k\pi
\end{array}$
+) TH2: $\dfrac{{\sin x - \cos x}}{{\sin x.\cos x}} = - 1(1)$
Ta đặt ${t = \sin x - \cos x\left( {\left| t \right| \le \sqrt 2 } \right)}$
$\begin{array}{l}
\Rightarrow {t^2} = {\sin ^2}x - 2\sin x.\cos x + {\cos ^2}x = 1 - 2\sin x.\cos x\\
\Rightarrow \sin x.\cos x = \dfrac{{1 - {t^2}}}{2}
\end{array}$
(1) trở thành:
$\begin{array}{l}
\dfrac{t}{{\dfrac{{1 - {t^2}}}{2}}} = - 1\\
\Leftrightarrow {t^2} - 2t - 1 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 1 + \sqrt 2(l) \\
t = 1 - \sqrt 2 (c)
\end{array} \right.
\end{array}$
Khi đó:
$\begin{array}{l}
\sin x - \cos x = 1 - \sqrt 2 \\
\Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = 1 - \dfrac{1}{{\sqrt 2 }}\\
\Leftrightarrow x + \dfrac{\pi }{4} = \pm \arccos \left( {1 - \dfrac{1}{{\sqrt 2 }}} \right) + k\pi (tm)
\end{array}$
Vậy phương trình có các nghiệm là:
$x = \dfrac{{3\pi }}{4} + k\pi \left( {k \in Z} \right)$; $x + \dfrac{\pi }{4} = \pm \arccos \left( {1 - \dfrac{1}{{\sqrt 2 }}} \right) + k\pi \left( {k \in Z} \right)$
