Toán Lớp 11: 4sin²x – 7cos2x + 9cosx – 11=0
sin5x – √3cos5x =2
Hix giúp mình vs ạ
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Toán Lớp 11: 4sin²x – 7cos2x + 9cosx – 11=0
sin5x – √3cos5x =2
Hix giúp mình vs ạ
Comments ( 1 )
a,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \pm \dfrac{\pi }{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{5}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
a,\\
4{\sin ^2}x – 7\cos 2x + 9\cos x – 11 = 0\\
\Leftrightarrow 4.\left( {1 – {{\cos }^2}x} \right) – 7.\left( {2{{\cos }^2}x – 1} \right) + 9\cos x – 11 = 0\\
\Leftrightarrow 4 – 4{\cos ^2}x – 14{\cos ^2}x + 7 + 9\cos x – 11 = 0\\
\Leftrightarrow – 18{\cos ^2}x + 9\cos x = 0\\
\Leftrightarrow – 9\cos x.\left( {2\cos x – 1} \right) = 0\\
\Leftrightarrow \cos x.\left( {2\cos x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
2\cos x – 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \pm \dfrac{\pi }{3} + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\sin 5x – \sqrt 3 \cos 5x = 2\\
\Leftrightarrow \dfrac{1}{2}\sin 5x – \dfrac{{\sqrt 3 }}{2}\cos 5x = 1\\
\Leftrightarrow \sin 5x.\cos \dfrac{\pi }{3} – \cos 5x.\sin \dfrac{\pi }{3} = 1\\
\Leftrightarrow \sin \left( {5x – \dfrac{\pi }{3}} \right) = 1\\
\Leftrightarrow 5x – \dfrac{\pi }{3} = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow 5x = \dfrac{{5\pi }}{6} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{5}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)