Toán Lớp 11: b, 2cos^2x – cosx – 1 = 0
c, sinx – căn 3 cos x = – căn 2
d, 5 sin^2x – sin2x – . cos^2x = 2
em xin lỗi vì ghi khó nhìn ạ, anh chị giúp em với ạ
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Toán Lớp 11: b, 2cos^2x – cosx – 1 = 0
c, sinx – căn 3 cos x = – căn 2
d, 5 sin^2x – sin2x – . cos^2x = 2
em xin lỗi vì ghi khó nhìn ạ, anh chị giúp em với ạ
Comments ( 1 )
b,\\
\left[ \begin{array}{l}
x = k2\pi \\
x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k2\pi \\
x = \dfrac{{19\pi }}{{12}} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
b,\\
2{\cos ^2}x – \cos x – 1 = 0\\
\Leftrightarrow \left( {2{{\cos }^2}x – 2\cos x} \right) + \left( {\cos x – 1} \right) = 0\\
\Leftrightarrow 2\cos x\left( {\cos x – 1} \right) + \left( {\cos x – 1} \right) = 0\\
\Leftrightarrow \left( {\cos x – 1} \right)\left( {2\cos x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x – 1 = 0\\
2\cos x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = – \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
\sin x – \sqrt 3 \cos x = – \sqrt 2 \\
\Leftrightarrow \dfrac{1}{2}\sin x – \dfrac{{\sqrt 3 }}{2}\cos x = – \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \sin x.\cos \dfrac{\pi }{3} – \cos x.\sin \dfrac{\pi }{3} = – \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \sin \left( {x – \dfrac{\pi }{3}} \right) = \sin \dfrac{{ – \pi }}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{\pi }{3} = – \dfrac{\pi }{4} + k2\pi \\
x – \dfrac{\pi }{3} = \pi – \dfrac{{ – \pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{\pi }{3} = – \dfrac{\pi }{4} + k2\pi \\
x – \dfrac{\pi }{3} = \dfrac{{5\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k2\pi \\
x = \dfrac{{19\pi }}{{12}} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)