Toán Lớp 11: 1. Cos ( 3x/5 – pi/3) = – căn 2/2
2. Cos ( x + 1) = 2/3
3. Sin (2x/3 – pi/6) = 1
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Toán Lớp 11: 1. Cos ( 3x/5 – pi/3) = – căn 2/2
2. Cos ( x + 1) = 2/3
3. Sin (2x/3 – pi/6) = 1
Giải hộ mik vs ạ . Mik cần gấp ạ
Cảm ơn nhiều
Comments ( 1 )
1)\cos \left( {\dfrac{{3x}}{5} – \dfrac{\pi }{3}} \right) = – \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{3x}}{5} – \dfrac{\pi }{3} = \dfrac{{3\pi }}{4} + k2\pi \\
\dfrac{{3x}}{5} – \dfrac{\pi }{3} = – \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{65\pi }}{{36}} + \dfrac{{k10\pi }}{3}\\
x = – \dfrac{{25\pi }}{{36}} + \dfrac{{k10\pi }}{3}
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x = \dfrac{{65\pi }}{{36}} + \dfrac{{k10\pi }}{3}\\
x = – \dfrac{{25\pi }}{{36}} + \dfrac{{k10\pi }}{3}
\end{array} \right.\\
2)\cos \left( {x + 1} \right) = \dfrac{2}{3}\\
\Leftrightarrow x + 1 = \pm \arccos \dfrac{2}{3} + k2\pi \\
\Leftrightarrow x = – 1 \pm \arccos \dfrac{2}{3} + k2\pi \\
Vậy\,x = – 1 \pm \arccos \dfrac{2}{3} + k2\pi \\
3)\sin \left( {\dfrac{{2x}}{3} – \dfrac{\pi }{6}} \right) = 1\\
\Leftrightarrow \dfrac{{2x}}{3} – \dfrac{\pi }{6} = \dfrac{\pi }{2} + 2k\pi \\
\Leftrightarrow x = \pi + 3k\pi \\
Vậy\,x = \pi + 3k\pi
\end{array}$