a) Cos2x-Sinx=1 b) √3 sinx-cosx=✓2

Question

a) Cos2x-Sinx=1
b) √3 sinx-cosx=✓2

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Calantha 5 years 2020-11-10T10:55:35+00:00 1 Answers 63 views 0

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  1. Eirian
    0
    2020-11-10T10:57:21+00:00
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    Đáp án:

    $\begin{array}{l}a) \,\left[\begin{array}{l}x = k\pi\\x =  -\dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} +k2\pi\end{array}\right. \quad (k \in \Bbb Z)\\b)\,\left[\begin{array}{l}x =\dfrac{5\pi}{12} +k2\pi\\x = \dfrac{11\pi}{12} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\end{array}$

    Giải thích các bước giải:

    $\begin{array}{l}a)\,\,\cos2x – \sin x = 1\\ \Leftrightarrow 1 – 2\sin^2x – \sin x = 1\\ \Leftrightarrow 2\sin^2x + \sin x = 0\\ \Leftrightarrow \sin x(2\sin x + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin x =0\\\sin x = – \dfrac{1}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\x = -\dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} +k2\pi\end{array}\right. \quad (k \in \Bbb Z)\\ b)\,\,\sqrt3\sin x – \cos x = \sqrt2\\ \Leftrightarrow \dfrac{\sqrt3}{2}\sin x – \dfrac{1}{2}\cos x = \dfrac{\sqrt2}{2}\\ \Leftrightarrow \sin x\cos\dfrac{\pi}{6} – \cos x\sin\dfrac{\pi}{6} = \sin\dfrac{\pi}{4}\\ \Leftrightarrow \sin\left(x – \dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{4}\\ \Leftrightarrow \left[\begin{array}{l}x – \dfrac{\pi}{6} = \dfrac{\pi}{4} +k2\pi\\x – \dfrac{\pi}{6} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =\dfrac{5\pi}{12} +k2\pi\\x = \dfrac{11\pi}{12} + k2\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$

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