1.Sin(2π/3 – 3x)= 0
2.sin(3x –π/6)=–1
3.sin(π/2 –x)=1
4.sin(x + 30°)=√3/2
5.sin(3x – π/4)=Cosx
Mn giúp e vs ạ
1.Sin(2π/3 – 3x)= 0
2.sin(3x –π/6)=–1
3.sin(π/2 –x)=1
4.sin(x + 30°)=√3/2
5.sin(3x – π/4)=Cosx
Mn giúp e vs ạ
Đáp án:
5) \(\left[ \begin{array}{l}
x = \dfrac{{3\pi }}{{16}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{3\pi }}{8} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\sin \left( { – 3x + \dfrac{{2\pi }}{3}} \right) = 0\\
\to – 3x + \dfrac{{2\pi }}{3} = k\pi \\
\to 3x = \dfrac{{2\pi }}{3} + k\pi \\
\to x = \dfrac{{2\pi }}{9} + \dfrac{{k\pi }}{3}\left( {k \in Z} \right)\\
2)\sin \left( {3x – \dfrac{\pi }{6}} \right) = – 1\\
\to 3x – \dfrac{\pi }{6} = – \dfrac{\pi }{2} + k2\pi \\
\to 3x = – \dfrac{\pi }{3} + k2\pi \\
\to x = – \dfrac{\pi }{9} + \dfrac{{k2\pi }}{3}\left( {k \in Z} \right)\\
3)\sin \left( { – x + \dfrac{\pi }{2}} \right) = 1\\
\to – x + \dfrac{\pi }{2} = \dfrac{\pi }{2} + k2\pi \\
\to x = k2\pi \left( {k \in Z} \right)\\
4)\sin \left( {x + \dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}\\
\to \left[ \begin{array}{l}
x + \dfrac{\pi }{6} = \dfrac{\pi }{3} + k2\pi \\
x + \dfrac{\pi }{6} = \pi – \dfrac{\pi }{3} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
5)\sin \left( {3x – \dfrac{\pi }{4}} \right) = \cos x\\
\to \sin \left( {3x – \dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{2} – x} \right)\\
\to \left[ \begin{array}{l}
3x – \dfrac{\pi }{4} = \dfrac{\pi }{2} – x + k2\pi \\
3x – \dfrac{\pi }{4} = \pi – \dfrac{\pi }{2} + x + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{3\pi }}{{16}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{3\pi }}{8} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
Đáp án:
Giải thích các bước giải: