2cos$^{2}$ (x+$\frac{\pi }{3}$ ) + 5sin(x + $\frac{\pi }{3}$ ) -4=0

Question

2cos$^{2}$ (x+$\frac{\pi }{3}$ ) + 5sin(x + $\frac{\pi }{3}$ ) -4=0

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Maria 2 tuần 2021-07-12T10:34:41+00:00 2 Answers 1 views 0

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    0
    2021-07-12T10:36:06+00:00

    Đáp án: `x=\frac{+π}{6}+k2π;x=\frac{π}{2}+k2π`

     

    Giải thích các bước giải:

     `2cos^2 (x+π/3) + 5sin (x+π/3) – 4 =0`

    `<=> 2(1-sin^2 (x+π/3)) + 5sin(x+π/3) – 4=0`

    `<=>  2sin^2 (x+π/3) * 5sin(x+π/3) + 2 =0` 

    Đặt `t= sin (x+π/3) ( -1 ≤t≤t)`, được:

    `2t^2 -5t+2=0 <=>` \(\left[ \begin{array}{l}t=2(L)\\t=\dfrac{1}{2}\end{array} \right.\) 

    Với `t = 1/2` có : `sin (x+π/3) = sin (π/6)`

    `<=>` \(\left[ \begin{array}{l}x+\dfrac{π}{3}=\dfrac{π}{6}+k2π\\x+\dfrac{π}{3}=π-\dfrac{π}{6}+k2π\end{array} \right.\) 

    `<=>` \(\left[ \begin{array}{l}x=\dfrac{-π}{6}+k2π\\x=\dfrac{π}{2}+k2π\end{array} \right.\) 

    Vậy PT có 2 họ nghiệm như trên.

    0
    2021-07-12T10:36:34+00:00

    `2cos² (x + π/3) + 5sin (x + π/3) – 4 = 0`

    Đặt `t = x + π/3`

    `=> 2cos² t + 5sin t – 4 = 0`

    `<=> 2 – 2sin² t + 5sin t – 4 = 0`

    `<=>` \(\left[ \begin{array}{l}sin t = 2 (l)\\sin t = \dfrac{1}{2}\end{array} \right.\) 

     `=> sin t = 1/2`

    `<=> sin (x + π/3) = sin (π/6)`

    `<=>` \(\left[ \begin{array}{l}x + \dfrac{π}{3} = \dfrac{π}{6} + k2π\\x + \dfrac{π}{3} = \dfrac{5π}{6} + k2π\end{array} \right.\) 

    `<=>` \(\left[ \begin{array}{l}x = -\dfrac{π}{6} + k2π\\x = \dfrac{π}{2} + k2π\end{array} \right.\) `(k ∈ ZZ)`

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