Đáp án:
\(\begin{array}{l}
1,\\
x = k\pi \,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
3,\\
\left[ \begin{array}{l}
x = - \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = \dfrac{1}{2}\arctan \dfrac{2}{3} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
2{\cos ^2}2x - 3{\sin ^2}x = 2\\
\Leftrightarrow 2.{\left( {1 - 2{{\sin }^2}x} \right)^2} - 3{\sin ^2}x - 2 = 0\\
\Leftrightarrow 2.\left[ {{1^2} - 2.1.2{{\sin }^2}x + {{\left( {2{{\sin }^2}x} \right)}^2}} \right] - 3{\sin ^2}x - 2 = 0\\
\Leftrightarrow 2.\left( {1 - 4{{\sin }^2}x + 4{{\sin }^4}x} \right) - 3{\sin ^2}x - 2 = 0\\
\Leftrightarrow 2 - 8{\sin ^2}x + 8{\sin ^4}x - 3{\sin ^2}x - 2 = 0\\
\Leftrightarrow 8{\sin ^4}x - 11{\sin ^2}x = 0\\
\Leftrightarrow {\sin ^2}x.\left( {8{{\sin }^2}x - 11} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}x = 0\\
8{\sin ^2}x - 11 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}x = 0\\
{\sin ^2}x = \dfrac{{11}}{8}
\end{array} \right.\\
- 1 \le \sin x \le 1 \Rightarrow 0 \le {\sin ^2}x \le 1\\
\Rightarrow {\sin ^2}x = 0\\
\Leftrightarrow \sin x = 0\\
\Leftrightarrow x = k\pi \,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
2 - {\cos ^2}x = {\sin ^4}x\\
\Leftrightarrow 2 - \left( {1 - {{\sin }^2}x} \right) = {\sin ^4}x\\
\Leftrightarrow 2 - 1 + {\sin ^2}x = {\sin ^4}x\\
\Leftrightarrow 1 + {\sin ^2}x = {\sin ^4}x\\
\Leftrightarrow {\sin ^4}x - {\sin ^2}x - 1 = 0\\
\Leftrightarrow \left( {{{\sin }^4}x - {{\sin }^2}x + \dfrac{1}{4}} \right) - \dfrac{5}{4} = 0\\
\Leftrightarrow \left[ {{{\left( {{{\sin }^2}x} \right)}^2} - 2.{{\sin }^2}x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] = \dfrac{5}{4}\\
\Leftrightarrow {\left( {{{\sin }^2}x - \dfrac{1}{2}} \right)^2} = \dfrac{5}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}x - \dfrac{1}{2} = \dfrac{{\sqrt 5 }}{2}\\
{\sin ^2}x - \dfrac{1}{2} = - \dfrac{{\sqrt 5 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}x = \dfrac{{\sqrt 5 + 1}}{2} > 1\\
{\sin ^2}x = \dfrac{{ - \sqrt 5 + 1}}{2} < 0
\end{array} \right.\\
- 1 \le \sin x \le 1 \Rightarrow 0 \le {\sin ^2}x \le 1\\
\Rightarrow Phương\,\,trình\,\,đã\,\,cho\,\,vô\,\,nghiệm\\
3,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
\sin 2x \ne 0\\
\cos 2x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x \ne k\pi \\
2x \ne \dfrac{\pi }{2} + k\pi
\end{array} \right. \Leftrightarrow 2x \ne \dfrac{{k\pi }}{2} \Leftrightarrow x \ne \dfrac{{k\pi }}{4}\\
3\tan 2x - 2\cot 2x + 1 = 0\\
\Leftrightarrow 3\tan 2x - 2.\dfrac{1}{{\tan 2x}} + 1 = 0\\
\Leftrightarrow 3.{\tan ^2}2x - 2 + \tan 2x = 0\\
\Leftrightarrow \left( {3{{\tan }^2}2x + 3\tan 2x} \right) + \left( { - 2\tan 2x - 2} \right) = 0\\
\Leftrightarrow 3\tan 2x\left( {\tan 2x + 1} \right) - 2.\left( {\tan 2x + 1} \right) = 0\\
\Leftrightarrow \left( {\tan 2x + 1} \right).\left( {3\tan 2x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\tan 2x + 1 = 0\\
3\tan 2x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\tan 2x = - 1\\
\tan 2x = \dfrac{2}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = - \dfrac{\pi }{4} + k\pi \\
2x = \arctan \dfrac{2}{3} + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = \dfrac{1}{2}\arctan \dfrac{2}{3} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
