Giải các phương trình sau: a) 2sin3x -1=0 b) 2sin(2x-π/4)-1=0 c) 2cos(2x-π/4)-1=0 d) 2tan(2x-π/4)-1=0 e) √3 cot(x-π/4)-1=0 ai giúp mình với ạ

Đáp án:

\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{24}} + k\pi \\
x = \dfrac{{13\pi }}{{24}} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
\left[ \begin{array}{l}
x = \dfrac{{7\pi }}{{24}} + k\pi \\
x =  - \dfrac{\pi }{{24}} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
d,\\
x = \dfrac{\pi }{8} + \dfrac{1}{2}\arctan \dfrac{1}{2} + \dfrac{{k\pi }}{2}\,\,\,\,\,\left( {k \in Z} \right)\\
e,\\
x = \dfrac{{7\pi }}{{12}} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
2\sin 3x - 1 = 0\\
 \Leftrightarrow \sin 3x = \dfrac{1}{2}\\
 \Leftrightarrow \sin 3x = \sin \dfrac{\pi }{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{6} + k2\pi \\
3x = \pi  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{6} + k2\pi \\
3x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
b,\\
2\sin \left( {2x - \dfrac{\pi }{4}} \right) - 1 = 0\\
 \Leftrightarrow \sin \left( {2x - \dfrac{\pi }{4}} \right) = \dfrac{1}{2}\\
 \Leftrightarrow \sin \left( {2x - \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{4} = \dfrac{\pi }{6} + k2\pi \\
2x - \dfrac{\pi }{4} = \pi  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{4} = \dfrac{\pi }{6} + k2\pi \\
2x - \dfrac{\pi }{4} = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{{5\pi }}{{12}} + k2\pi \\
2x = \dfrac{{13\pi }}{{12}} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{24}} + k\pi \\
x = \dfrac{{13\pi }}{{24}} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
2\cos \left( {2x - \dfrac{\pi }{4}} \right) - 1 = 0\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{4}} \right) = \dfrac{1}{2}\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{4}} \right) = \cos \dfrac{\pi }{3}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{4} = \dfrac{\pi }{3} + k2\pi \\
2x - \dfrac{\pi }{4} =  - \dfrac{\pi }{3} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{{7\pi }}{{12}} + k2\pi \\
2x =  - \dfrac{\pi }{{12}} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{7\pi }}{{24}} + k\pi \\
x =  - \dfrac{\pi }{{24}} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
d,\\
DKXD:\,\,\,\cos \left( {2x - \dfrac{\pi }{4}} \right) \ne 0\\
 \Leftrightarrow 2x - \dfrac{\pi }{4} \ne \dfrac{\pi }{2} + k\pi  \Leftrightarrow x \ne \dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)\\
2\tan \left( {2x - \dfrac{\pi }{4}} \right) - 1 = 0\\
 \Leftrightarrow \tan \left( {2x - \dfrac{\pi }{4}} \right) = \dfrac{1}{2}\\
 \Leftrightarrow 2x - \dfrac{\pi }{4} = \arctan \dfrac{1}{2} + k\pi \\
 \Leftrightarrow 2x = \dfrac{\pi }{4} + \arctan \dfrac{1}{2} + k\pi \\
 \Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{1}{2}\arctan \dfrac{1}{2} + \dfrac{{k\pi }}{2}\,\,\,\,\,\left( {k \in Z} \right)\\
e,\\
DKXD:\,\,\sin \left( {x - \dfrac{\pi }{4}} \right) \ne 0 \Leftrightarrow x - \dfrac{\pi }{4} \ne k\pi  \Leftrightarrow x \ne \dfrac{\pi }{4} + k\pi \,\,\,\left( {k \in Z} \right)\\
\sqrt 3 \cot \left( {x - \dfrac{\pi }{4}} \right) - 1 = 0\\
 \Leftrightarrow \cot \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 3 }}\\
 \Leftrightarrow x - \dfrac{\pi }{4} = \dfrac{\pi }{3} + k\pi \\
 \Leftrightarrow x = \dfrac{{7\pi }}{{12}} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)