Giải pt : cos( 2x + 2π/ 3 ) + 3.cos( x + π/3 ) +1 = 0

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Giải pt : cos( 2x + 2π/ 3 ) + 3.cos( x + π/3 ) +1 = 0

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Ladonna 4 years 2020-10-25T11:04:32+00:00 1 Answers 83 views 0

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  1. Philomena
    0
    2020-10-25T11:06:11+00:00
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    Đáp án:

    $x = \dfrac{\pi}{6} +k\pi \quad (k \in \Bbb Z)$

    Giải thích các bước giải:

    $\begin{array}{l}\cos\left(2x + \dfrac{2\pi}{3}\right) + 3\cos\left(x + \dfrac{\pi}{3}\right) + 1 = 0\\ \Leftrightarrow \cos\left[2\left(x + \dfrac{\pi}{3}\right)\right] + 3\cos\left(x + \dfrac{\pi}{3}\right) + 1 = 0\\ \Leftrightarrow 2\cos^2\left(x + \dfrac{\pi}{3}\right) – 1 + 3\cos\left(x + \dfrac{\pi}{3}\right) + 1 = 0\\ \Leftrightarrow \cos\left(x + \dfrac{\pi}{3}\right)\left[2\cos\left(x + \dfrac{\pi}{3}\right) + 3 \right] = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos\left(x + \dfrac{\pi}{3}\right) = 0\\\cos\left(x + \dfrac{\pi}{3}\right) = – \dfrac{3}{2} \quad (loại)\end{array}\right.\\ \Leftrightarrow x + \dfrac{\pi}{3} = \dfrac{\pi}{2} + k\pi\\ \Leftrightarrow x = \dfrac{\pi}{6} +k\pi \quad (k \in \Bbb Z)\end{array}$

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