Mn giúp em nhanh với ạ em đang xần rất gấp .cảm ơn mn nhiều Question Mn giúp em nhanh với ạ em đang xần rất gấp .cảm ơn mn nhiều in progress 0 Môn Toán Hưng Gia 4 years 2020-10-17T03:17:09+00:00 2020-10-17T03:17:09+00:00 1 Answers 84 views 0
Answers ( )
Đáp án:
Vậy phương trình có các họ nghiệm là:
$a)x = \dfrac{\pi }{4} + k\dfrac{\pi }{2};x = \dfrac{\pi }{{12}} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$
$b)x = \dfrac{\pi }{4} + k\pi ;x = \arctan \left( { – 5} \right) + k\pi \left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
a)2{\cos ^2}2x – 3\sqrt 3 \sin 4x – 4{\sin ^2}2x = – 4\\
\Leftrightarrow 2{\cos ^2}2x – 6\sqrt 3 \sin 2x\cos 2x – 4{\sin ^2}2x = – 4{\cos ^2}2x – 4{\sin ^2}2x\\
\Leftrightarrow 6{\cos ^2}2x – 6\sqrt 3 \sin 2x\cos 2x = 0\\
\Leftrightarrow 6\cos 2x\left( {\cos 2x – \sqrt 3 \sin 2x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
\cos 2x – \sqrt 3 \sin 2x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
\dfrac{1}{2}\cos 2x – \dfrac{{\sqrt 3 }}{2}\sin 2x = 0
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
\cos \left( {2x + \dfrac{\pi }{3}} \right) = 0
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\\
2x + \dfrac{\pi }{3} = \dfrac{\pi }{2} + k\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\\
x = \dfrac{\pi }{{12}} + k\dfrac{\pi }{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
$\begin{array}{l}
b){\sin ^2}x + \sin 2x – 2{\cos ^2}x = \dfrac{1}{2}\\
\Leftrightarrow 2{\sin ^2}x + 2\sin 2x – 4{\cos ^2}x = 1\\
\Leftrightarrow 2{\sin ^2}x + 4\sin x\cos x – 4{\cos ^2}x = {\sin ^2}x + {\cos ^2}x\\
\Leftrightarrow {\sin ^2}x + 4\sin x\cos x – 5{\cos ^2}x = 0\\
\Leftrightarrow \left( {\sin x – \cos x} \right)\left( {\sin x + 5\cos x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x – \cos x = 0\\
\sin x + 5\cos x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = \cos x\\
\sin x = – 5\cos x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = – 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \arctan \left( { – 5} \right) + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$