2. 3 tan ( x/2 -pi/3)= căn 3 3. căn 3 Sin x + Cos x = 2 Giúp mik vs cần gấp cảm ơn mọi ng nhìu

Question

2. 3 tan ( x/2 -pi/3)= căn 3
3. căn 3 Sin x + Cos x = 2

Giúp mik vs cần gấp cảm ơn mọi ng nhìu
2-3-tan-2-pi-3-can-3-3-can-3-sin-cos-2-giup-mik-vs-can-gap-cam-on-moi-ng-nhiu

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Nho 6 years 2020-11-21T15:50:34+00:00 2 Answers 95 views 0

Answers ( )

  1. Bo
    0
    2020-11-21T15:52:14+00:00
    Reply

    Đáp án:

    $2) \, x = \pi + k2\pi \quad (k \in \Bbb Z)$

    $3) \, x = \dfrac{\pi}{3} + k2\pi\quad (k \in \Bbb Z)$

    Giải thích các bước giải:

    $2) \,\,3\tan\left(\dfrac{x}{2} – \dfrac{\pi}{3}\right) = \sqrt3\qquad (*)$

    $ĐKXĐ: \, \cos\left(\dfrac{x}{2} – \dfrac{\pi}{3}\right) \ne 0 \Leftrightarrow \dfrac{5\pi}{3} + n2\pi \quad (n \in \Bbb Z)$

    $(*)\Leftrightarrow \tan\left(\dfrac{x}{2} – \dfrac{\pi}{3}\right) = \dfrac{\sqrt3}{3}$

    $\Leftrightarrow \tan\left(\dfrac{x}{2} – \dfrac{\pi}{3}\right) = \tan\dfrac{\pi}{6}$

    $\Leftrightarrow \dfrac{x}{2} – \dfrac{\pi}{3} = \dfrac{\pi}{6} + k\pi$

    $\Leftrightarrow \dfrac{x}{2} = \dfrac{\pi}{2} + k\pi$

    $\Leftrightarrow x = \pi + k2\pi \quad (k \in \Bbb Z)$

    $3)\,\, \sqrt3\sin x + \cos x = 2$

    $\Leftrightarrow \dfrac{\sqrt3}{2}\sin x + \dfrac{1}{2}\cos x = 1$

    $\Leftrightarrow \sin x\cos\dfrac{\pi}{6} + \cos x\sin\dfrac{\pi}{6} = 1$

    $\Leftrightarrow \sin\left(x + \dfrac{\pi}{6}\right) = 1$

    $\Leftrightarrow x + \dfrac{\pi}{6} = \dfrac{\pi}{2} + k\pi$

    $\Leftrightarrow x = \dfrac{\pi}{3} + k2\pi\quad (k \in \Bbb Z)$

  2. Dau
    0
    2020-11-21T15:52:32+00:00
    Reply

    Giải thích các bước giải:

     $\text{2) 3tan($\dfrac{x}{2}$ – $\dfrac{\pi}{3}$ )=$\sqrt[]{3}$ }$

    ⇔$\text{tan($\dfrac{x}{2}$ – $\dfrac{\pi}{3}$ )=$\dfrac{\sqrt[]{3}}{3}$ }$

    ⇔$\text{$\dfrac{x}{2}$ – $\dfrac{\pi}{3}$ = $\dfrac{\pi}{6}$ +k$\pi$}$

    ⇔$\text{$\dfrac{x}{2}$ = $\dfrac{\pi}{2}$ +k$\pi$}$

    ⇒$\text{${x}$ = $\pi$ +k2$\pi$ (k∈Z)}$

     $\text{3) $\sqrt[]{3}$sinx + cosx = 2 }$

      chia hai vế cho 2 ta được:

    ⇒$\text{ $\dfrac{\sqrt[]{3}}{2}$ sinx + $\dfrac{1}{2}$ cosx = 1 }$

    ⇔$\text{ sin(x + $\dfrac{\pi}{6}$)=1 }$

    ⇔$\text{ x + $\dfrac{\pi}{6}$= $\dfrac{\pi}{2}$ +k2$\pi$ }$

    ⇒$\text{ x = $\dfrac{\pi}{3}$ +k2$\pi$ (k∈Z) }$

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