Toán Lớp 11: giải gấp giúp em chi tiết 3 bài này với ạ
a) sin4x +cos4x = √3
b) 3sinx + √3.cosx = 1
c) √3.cosx + sinx = – √2
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Toán Lớp 11: giải gấp giúp em chi tiết 3 bài này với ạ
a) sin4x +cos4x = √3
b) 3sinx + √3.cosx = 1
c) √3.cosx + sinx = – √2
Comments ( 2 )
a)\sin 4x + \cos 4x = \sqrt 3 \\
\Leftrightarrow \sqrt 2 .\sin \left( {4x + \dfrac{\pi }{4}} \right) = \sqrt 3 \\
\Leftrightarrow \sin \left( {4x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 3 }}{{\sqrt 2 }}\\
Do:\left\{ \begin{array}{l}
– 1 \le \sin \left( {4x + \dfrac{\pi }{4}} \right) \le 1\\
\sqrt {\dfrac{3}{2}} > 1
\end{array} \right.
\end{array}$
b)3\sin x + \sqrt 3 .\cos x = 1\\
\Leftrightarrow \dfrac{3}{{2\sqrt 3 }}\sin x + \dfrac{{\sqrt 3 }}{{2\sqrt 3 }}\cos x = \dfrac{1}{{2\sqrt 3 }}\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin x + \dfrac{1}{2}\cos x = \dfrac{1}{{2\sqrt 3 }}\\
\Leftrightarrow \sin \dfrac{\pi }{3}.\sin x + \cos \dfrac{\pi }{3}.\cos x = \dfrac{1}{{2\sqrt 3 }}\\
\Leftrightarrow \cos \left( {x – \dfrac{\pi }{3}} \right) = \dfrac{1}{{2\sqrt 3 }}\\
\Leftrightarrow x – \dfrac{\pi }{3} = \pm \arccos \dfrac{1}{{2\sqrt 3 }} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{3} \pm \arccos \dfrac{1}{{2\sqrt 3 }} + k2\pi \\
Vậy\,x = \dfrac{\pi }{3} \pm \arccos \dfrac{1}{{2\sqrt 3 }} + k2\pi \\
\end{array}$
c)\sqrt 3 \cos x + \sin x = – \sqrt 2 \\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos x + \dfrac{1}{2}\sin x = – \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \cos \left( {x – \dfrac{\pi }{6}} \right) = \dfrac{{ – \sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{\pi }{6} = \dfrac{{3\pi }}{4} + k2\pi \\
x – \dfrac{\pi }{6} = – \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{11\pi }}{{12}} + k2\pi \\
x = \dfrac{{ – 7\pi }}{{12}} + k2\pi
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x = \dfrac{{11\pi }}{{12}} + k2\pi \\
x = \dfrac{{ – 7\pi }}{{12}} + k2\pi
\end{array} \right.
\end{array}$