Toán Lớp 11: a) sin x – cos x = \sqrt{2}
b) 2sin2x + 2cos2x = \sqrt{-2}
c) cos x – \sqrt{3} sinx = $\sqrt{3}$
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Toán Lớp 11: a) sin x – cos x = \sqrt{2}
b) 2sin2x + 2cos2x = \sqrt{-2}
c) cos x – \sqrt{3} sinx = $\sqrt{3}$
Comments ( 2 )
a)\quad \sin x – \cos x = \sqrt2\\
\Leftrightarrow \sqrt2\sin\left(x – \dfrac{\pi}{4}\right)= \sqrt2\\
\Leftrightarrow \sin\left(x – \dfrac{\pi}{4}\right)= 1\\
\Leftrightarrow x – \dfrac{\pi}{4} = \dfrac{\pi}{2} + k2\pi\\
\Leftrightarrow x = \dfrac{3\pi}{4} +k2\pi\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{3\pi}{4} +k2\pi\ \Bigg|\ k\in\Bbb Z\right\}\\
b)\quad \text{Sửa đề:}\ \ 2\sin2x + 2\cos2x= – \sqrt2\\
\Leftrightarrow \sin2x + \cos2x = – \dfrac{\sqrt2}{2}\\
\Leftrightarrow \sqrt2\sin\left(2x + \dfrac{\pi}{4}\right) = – \dfrac{\sqrt2}{2}\\
\Leftrightarrow \sin\left(2x + \dfrac{\pi}{4}\right) = – \dfrac12\\
\Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{4} = – \dfrac{\pi}{6} +k2\pi\\2x + \dfrac{\pi}{4} = \dfrac{7\pi}{6} +k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = – \dfrac{5\pi}{24} +k\pi\\x = \dfrac{11\pi}{24} + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{- \dfrac{5\pi}{24} +k\pi;\ \dfrac{11\pi}{24} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\\
c)\quad \cos x – \sqrt3\sin x = \sqrt3\\
\Leftrightarrow \dfrac12\cos x – \dfrac{\sqrt3}{2}\sin x = \dfrac{\sqrt3}{2}\\
\Leftrightarrow \cos\left(x + \dfrac{\pi}{3}\right) = \cos\dfrac{\pi}{6}\\
\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{3} = – \dfrac{\pi}{6} +k2\pi\\x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{2} +k2\pi\\x =- \dfrac{\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{- \dfrac{\pi}{2} +k2\pi;\ – \dfrac{\pi}{6} + k2\pi\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)