Giải giúp mình câu q vs ạ

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Giải giúp mình câu q vs ạ
giai-giup-minh-cau-q-vs-a

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Khải Quang 4 years 2020-11-04T15:45:41+00:00 2 Answers 62 views 0

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    0
    2020-11-04T15:46:55+00:00

    Đáp án:

    $\left[\begin{array}{l}x \approx \pm \arccos(-0,6462) + k2\pi\\x\approx \pm \arccos0,19865 + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$

    Giải thích các bước giải:

    $\begin{array}{l}\dfrac{1}{\cos x} = 4\sin^2x + 6\cos x \qquad (*)\\ ĐKXĐ: \, \cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi}{2} + n\pi \quad (n \in \Bbb Z)\\ (*) \Leftrightarrow 4(1 – \cos^2x).\cos x + 6\cos^2x – 1 = 0\\ \Leftrightarrow 4\cos^3x – 6\cos^2x – 4\cos x + 1 = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos x \approx -0,6462\\\cos x\approx 0,19865\\\cos x \approx 1,9476 \quad (loại)\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x \approx \pm \arccos(-0,6462) + k2\pi\\x\approx \pm \arccos0,19865 + k2\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$

    0
    2020-11-04T15:47:00+00:00

    `ĐK: x ne π/2 + kπ`

    `PT`

    `=> 1 = 4(1 – cos² x).cos x + 6cos² x`

    `<=> 1 = 4cos x – 4cos³ x + 6cos² x`

    `<=>` \(\left[ \begin{array}{l}cos x ≈ 1,95 (l)\\cos x ≈ 0,2\\cos x ≈ -0,65\end{array} \right.\) 

    `<=>` \(\left[ \begin{array}{l}x ≈ ±arccos 0,2 + k2π\\x ≈ ±arccos -0,65 + k2π\end{array} \right.\) `(k ∈ ZZ)`

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