giải pt sau giải từng bước giúp e ạ November 15, 2020 by Calantha giải pt sau giải từng bước giúp e ạ
Giải thích các bước giải: Ta có: \(\begin{array}{l}e,\\2{\cos ^2}2x = 1\\ \Leftrightarrow {\cos ^2}2x = \dfrac{1}{2}\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \dfrac{{\sqrt 2 }}{2}\\\cos 2x = – \dfrac{{\sqrt 2 }}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2x = \pm \dfrac{\pi }{4} + k2\pi \\2x = \pm \dfrac{{3\pi }}{4} + k2\pi \end{array} \right.\\ \Leftrightarrow 2x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\ \Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\\f,\\4{\cos ^2}4x = 1\\ \Leftrightarrow {\cos ^2}4x = \dfrac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos 4x = \dfrac{1}{2}\\\cos 4x = – \dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}4x = \pm \dfrac{\pi }{3} + k2\pi \\4x = \pm \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \pm \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\\x = \pm \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\end{array} \right.\\g,\\4{\cos ^2}5x = 3\\ \Leftrightarrow {\cos ^2}5x = \dfrac{3}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos 5x = \dfrac{{\sqrt 3 }}{2}\\\cos 5x = – \dfrac{{\sqrt 3 }}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}5x = \pm \dfrac{\pi }{6} + k2\pi \\5x = \pm \dfrac{{5\pi }}{6} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \pm \dfrac{\pi }{{30}} + \dfrac{{k2\pi }}{5}\\x = \pm \dfrac{\pi }{6} + \dfrac{{k2\pi }}{5}\end{array} \right.\\h,\\8{\cos ^2}x = 3\\ \Leftrightarrow {\cos ^2}x = \dfrac{3}{8}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \dfrac{{\sqrt 6 }}{4}\\\cos x = – \dfrac{{\sqrt 6 }}{4}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \pm \arccos \dfrac{{\sqrt 6 }}{4} + k2\pi \\x = \pm \arccos \left( { – \dfrac{{\sqrt 6 }}{4}} \right) + k2\pi \end{array} \right.\end{array}\) Reply
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
e,\\
2{\cos ^2}2x = 1\\
\Leftrightarrow {\cos ^2}2x = \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = \dfrac{{\sqrt 2 }}{2}\\
\cos 2x = – \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \pm \dfrac{\pi }{4} + k2\pi \\
2x = \pm \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow 2x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
\Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\\
f,\\
4{\cos ^2}4x = 1\\
\Leftrightarrow {\cos ^2}4x = \dfrac{1}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 4x = \dfrac{1}{2}\\
\cos 4x = – \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4x = \pm \dfrac{\pi }{3} + k2\pi \\
4x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\\
x = \pm \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}
\end{array} \right.\\
g,\\
4{\cos ^2}5x = 3\\
\Leftrightarrow {\cos ^2}5x = \dfrac{3}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 5x = \dfrac{{\sqrt 3 }}{2}\\
\cos 5x = – \dfrac{{\sqrt 3 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
5x = \pm \dfrac{\pi }{6} + k2\pi \\
5x = \pm \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{\pi }{{30}} + \dfrac{{k2\pi }}{5}\\
x = \pm \dfrac{\pi }{6} + \dfrac{{k2\pi }}{5}
\end{array} \right.\\
h,\\
8{\cos ^2}x = 3\\
\Leftrightarrow {\cos ^2}x = \dfrac{3}{8}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{\sqrt 6 }}{4}\\
\cos x = – \dfrac{{\sqrt 6 }}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \arccos \dfrac{{\sqrt 6 }}{4} + k2\pi \\
x = \pm \arccos \left( { – \dfrac{{\sqrt 6 }}{4}} \right) + k2\pi
\end{array} \right.
\end{array}\)