Toán Lớp 11: Giải phương trình:
$\dfrac{1+sinx+cos2x)sin(x+\dfrac{π}{4})}{1+tanx}=\dfrac{1}{\sqrt{2}}cos x$
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Toán Lớp 11: Giải phương trình:
$\dfrac{1+sinx+cos2x)sin(x+\dfrac{π}{4})}{1+tanx}=\dfrac{1}{\sqrt{2}}cos x$
Comments ( 2 )
Nè bạn ơi vote cho mk
$\begin{array}{l}
\dfrac{{\left( {1 + \sin x + \cos 2x} \right)\sin \left( {x + \dfrac{\pi }{4}} \right)}}{{1 + \tan x}} = \dfrac{1}{{\sqrt 2 }}\cos x\\
\text{ĐK}:\left\{ \begin{array}{l}
\cos x \ne 0\\
1 + \tan x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{2} + k\pi \\
x \ne – \dfrac{\pi }{4} + k\pi
\end{array} \right.\\
\Leftrightarrow \dfrac{{\left( {1 + \sin x + \cos 2x} \right)\dfrac{1}{{\sqrt 2 }}\left( {\sin x + \cos x} \right)}}{{1 + \tan x}} = \dfrac{1}{{\sqrt 2 }}\cos x\\
\Leftrightarrow \dfrac{{\left( {1 + \sin x + \cos 2x} \right)\left( {\sin x + \cos x} \right)}}{{1 + \tan x}} = \cos x\\
\Leftrightarrow \left( {1 + \sin x + \cos 2x} \right)\left( {\sin x + \cos x} \right) = \cos x\left( {1 + \tan x} \right)\\
\Leftrightarrow \left( {1 + \sin x + \cos 2x} \right)\left( {\sin x + \cos x} \right) = \cos x + \sin x\\
\Leftrightarrow \left( {\sin x + \cos x} \right)\left( {\sin x + \cos 2x + 1 – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x + \cos x = 0\\
\sin x + \cos 2x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\
– 2{\sin ^2}x + \sin x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\
\sin x = 1\\
\sin x = – \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{\pi }{4} + k\pi (L)\\
x = \dfrac{\pi }{2} + k2\pi (L)\\
x = – \dfrac{\pi }{6} + k2\pi ™\\
x = \dfrac{{7\pi }}{6} + k2\pi ™
\end{array} \right.\\
\Rightarrow S = \left\{ { – \dfrac{\pi }{6} + k2\pi ;\dfrac{{7\pi }}{6} + k2\pi ,k \in Z} \right\}
\end{array}$