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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: tan(-x)=tan(2x+1), x ∈[- $\pi$ ; $\pi$ ]

Home/ Toán Lớp 11: tan(-x)=tan(2x+1), x ∈[- $\pi$ ; $\pi$ ]

Toán Lớp 11: tan(-x)=tan(2x+1), x ∈[- $\pi$ ; $\pi$ ]

Tháng Một 6, 2023 Toán học 2 Comments 0 views

Toán Lớp 11: tan(-x)=tan(2x+1), x ∈[- $\pi$ ; $\pi$ ]

Comments ( 2 )

  1. ngoclankhue
    ngoclankhue
    6 Tháng Một, 2023 at 9:45 sáng
    Reply
    Giải đáp:
    $k= -3 -> x=\dfrac{-1}{3}+\pi$
    $k=-2 -> x=\dfrac{-1}{3}+\dfrac{2\pi}{3}$
    $k=-1 -> x=\dfrac{-1}{3}+\dfrac{\pi}{3}$
    $k=0->x=\dfrac{-1}{3}$
    $k=1-> x=\dfrac{-1}{3}-\dfrac{\pi}{3}$
    $k=2 -> x=\dfrac{-1}{3}-\dfrac{2\pi}{3}$
    $k= 3 -> x=\dfrac{-1}{3}-\pi$
     
    Lời giải và giải thích chi tiết:
     $Tan(-x)=tan(2x+1)$
    $\Leftrightarrow -x= 2x+1+k\pi$
    $\Leftrightarrow x=\dfrac{-1}{3}-\dfrac{k\pi}{3}$
    Để $x\in [-\pi ;\pi]$ thì
    $-\pi \leq \dfrac{-1}{3}-\dfrac{k\pi}{3} \leq \pi$
    $\dfrac{-1}{3}-\pi\leq \dfrac{k\pi}{3} \leq \pi +\dfrac{1}{3}$
    $\dfrac{-1-3\pi}{3}\leq \dfrac{k\pi}{3} \leq \dfrac{1+3\pi}{3}$
    $-1-3\pi \leq k\pi \leq 1+3\pi$
    $-3-\dfrac{1}{\pi} \leq k \leq 3+\dfrac{1}{\pi}$
    $k= \{\ -3;-2;-1; 0;1;2;3 \}$
    Vậy với :
    $k= -3 -> x=\dfrac{-1}{3}+\pi$
    $k=-2 -> x=\dfrac{-1}{3}+\dfrac{2\pi}{3}$
    $k=-1 -> x=\dfrac{-1}{3}+\dfrac{\pi}{3}$
    $k=0->x=\dfrac{-1}{3}$
    $k=1-> x=\dfrac{-1}{3}-\dfrac{\pi}{3}$
    $k=2 -> x=\dfrac{-1}{3}-\dfrac{2\pi}{3}$
    $k= 3 -> x=\dfrac{-1}{3}-\pi$

  2. truongankimtrong
    truongankimtrong
    6 Tháng Một, 2023 at 9:45 sáng
    Reply
    $\begin{array}{l} \tan \left( { – x} \right) = \tan \left( {2x + 1} \right)\\  \Leftrightarrow  – x = 2x + 1 + k\pi \\  \Leftrightarrow 3x =  – 1 – k\pi \\  \Leftrightarrow x = \dfrac{{ – 1}}{3} – \dfrac{{k\pi }}{3}\left( {k \in Z} \right)\\ x \in \left[ { – \pi ;\pi } \right] \Rightarrow  – \pi  \le \dfrac{{ – 1}}{3} – \dfrac{{k\pi }}{3} \le \pi \\  \Leftrightarrow  – 1 \le \dfrac{{ – 1}}{{3\pi }} – \dfrac{k}{3} \le 1\\  \Leftrightarrow   \dfrac{1}{{3\pi }} – 1 \le \dfrac{k}{3} \le \dfrac{1}{{3\pi }} + 1\\  \Leftrightarrow k \in \left\{ { -3;-2;- 1;0;1;2;3} \right\} \end{array}$
    $\begin{array}{l}
    k =  – 3 \Rightarrow x = \dfrac{{ – 1}}{3} + \pi \\
    k =  – 2 \Rightarrow x = \dfrac{{ – 1}}{3} + \dfrac{{2\pi }}{3}\\
    k =  – 1 \Rightarrow x = \dfrac{{ – 1}}{3} + \dfrac{\pi }{3}\\
    k = 0 \Rightarrow x =  – \dfrac{1}{3}\\
    k = 1 \Rightarrow x = \dfrac{{ – 1}}{3} – \dfrac{\pi }{3}\\
    k = 2 \Rightarrow x = \dfrac{{ – 1}}{3} – \dfrac{{2\pi }}{3}\\
    k = 3 \Rightarrow x =  – \dfrac{1}{3} – \pi 
    \end{array}$

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222-9+11+12:2*14+14 = ? ( )

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