Toán Lớp 11: nghiệm của phương trình: 2(sin^4 x+ cos^4 x)+ √3 sin4x =2
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: nghiệm của phương trình: 2(sin^4 x+ cos^4 x)+ √3 sin4x =2
Comments ( 1 )
x = \dfrac{{k\pi }}{2}\\
x = \dfrac{1}{2}\arctan 2\sqrt 3 + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\)
2\left( {{{\sin }^4}x + {{\cos }^4}x} \right) + \sqrt 3 \sin 4x = 2\\
\Leftrightarrow 2.\left[ {\left( {{{\sin }^4}x + 2{{\sin }^2}x.{{\cos }^2}x + {{\cos }^4}x} \right) – 2{{\sin }^2}x.{{\cos }^2}x} \right] + \sqrt 3 \sin 4x = 2\\
\Leftrightarrow 2.\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} – 2.{{\sin }^2}x.{{\cos }^2}x} \right] + \sqrt 3 \sin 4x = 2\\
\Leftrightarrow 2.\left( {{1^2} – 2{{\sin }^2}x.{{\cos }^2}x} \right) + \sqrt 3 \sin 4x – 2 = 0\\
\Leftrightarrow 2 – 4{\sin ^2}x.{\cos ^2}x + \sqrt 3 \sin 4x – 2 = 0\\
\Leftrightarrow 4{\sin ^2}x.{\cos ^2}x – \sqrt 3 .\sin 4x = 0\\
\Leftrightarrow {\left( {2\sin x.\cos x} \right)^2} – \sqrt 3 \sin 4x = 0\\
\Leftrightarrow {\left( {\sin 2x} \right)^2} – \sqrt 3 .2\sin 2x.\cos 2x = 0\\
\Leftrightarrow {\sin ^2}2x – 2\sqrt 3 \sin 2x.\cos 2x = 0\\
\Leftrightarrow \sin 2x.\left( {\sin 2x – 2\sqrt 3 \cos 2x} \right) = 0\,\,\,\,\left( 1 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\sin 2x – 2\sqrt 3 \cos 2x = 0
\end{array} \right.\,\,\,\,\,\left( * \right)\\
TH1:\,\,\,\cos 2x = 0\\
\left( 1 \right) \Leftrightarrow \sin 2x.\sin 2x = 0 \Leftrightarrow \sin 2x = 0\\
\Rightarrow {\sin ^2}2x + {\cos ^2}2x = 0\,\,\,\,\left( L \right)\\
TH2:\,\,\,\cos 2x \ne 0\\
\left( * \right) \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\dfrac{{\sin 2x}}{{\cos 2x}} – 2\sqrt 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = k\pi \\
\tan 2x = 2\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
2x = \arctan 2\sqrt 3 + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x = \dfrac{1}{2}\arctan 2\sqrt 3 + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)