Đáp án:
\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = \arctan \dfrac{5}{2} + k\pi \\
x = - \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{12}} + k2\pi \\
x = \dfrac{{13\pi }}{{12}} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
2{\sin ^2}x - 3\sin x.\cos x - 5{\cos ^2}x = 0\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\cos x = 0\\
\left( 1 \right) \Leftrightarrow 2{\sin ^2}x = 0 \Leftrightarrow \sin x = 0\\
\Rightarrow {\sin ^2}x + {\cos ^2}x = 0\,\,\,\,\left( L \right)\\
TH2:\,\,\,\,\cos x \ne 0\\
\left( 1 \right) \Leftrightarrow \dfrac{{2{{\sin }^2}x - 3\sin x.\cos x - 5{{\cos }^2}x}}{{{{\cos }^2}x}} = 0\\
\Leftrightarrow 2.\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - 3.\dfrac{{\sin x}}{{\cos x}} - 5 = 0\\
\Leftrightarrow 2.{\tan ^2}x - 3\tan x - 5 = 0\\
\Leftrightarrow \left( {2{{\tan }^2}x - 5\tan x} \right) + \left( {2\tan x - 5} \right) = 0\\
\Leftrightarrow \tan x.\left( {2\tan x - 5} \right) + \left( {2\tan x - 5} \right) = 0\\
\Leftrightarrow \left( {2\tan x - 5} \right).\left( {\tan x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2\tan x - 5 = 0\\
\tan x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = \dfrac{5}{2}\\
\tan x = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \arctan \dfrac{5}{2} + k\pi \\
x = - \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\sqrt 2 \sin x - \sqrt 2 \cos x = 1\\
\Leftrightarrow \dfrac{{\sqrt 2 }}{2}\sin x - \dfrac{{\sqrt 2 }}{2}\cos x = \dfrac{1}{2}\\
\Leftrightarrow \sin x.\cos \dfrac{\pi }{4} - \cos x.\sin \dfrac{\pi }{4} = \dfrac{1}{2}\\
\Leftrightarrow \sin \left( {x - \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{4} = \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{\pi }{4} = \pi - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{12}} + k2\pi \\
x = \dfrac{{13\pi }}{{12}} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
