Toán Lớp 11: 2cos2x -3sin²x=1
Giải pt theo công thức hạ bậc. Giúp mik ạ
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 11: 2cos2x -3sin²x=1
Giải pt theo công thức hạ bậc. Giúp mik ạ
Comments ( 1 )
x = \pm \arccos \dfrac{{\sqrt {42} }}{7} + k2\pi \\
x = \pm \arccos \dfrac{{ – \sqrt {42} }}{7} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
2\cos 2x – 3{\sin ^2}x = 1\\
\Leftrightarrow 2.\left( {1 – 2{{\sin }^2}x} \right) – 3{\sin ^2}x = 1\\
\Leftrightarrow 2 – 4{\sin ^2}x – 3{\sin ^2}x – 1 = 0\\
\Leftrightarrow 1 – 7{\sin ^2}x = 0\\
\Leftrightarrow {\sin ^2}x = \dfrac{1}{7}\\
\Rightarrow {\cos ^2}x = 1 – {\sin ^2}x = \dfrac{6}{7}\\
\Rightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{\sqrt {42} }}{7}\\
\cos x = – \dfrac{{\sqrt {42} }}{7}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \arccos \dfrac{{\sqrt {42} }}{7} + k2\pi \\
x = \pm \arccos \dfrac{{ – \sqrt {42} }}{7} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)