giải từng bước giúp e 5 bài này với ạ thanks nhìu ạ

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giải từng bước giúp e 5 bài này với ạ thanks nhìu ạ
giai-tung-buoc-giup-e-5-bai-nay-voi-a-thanks-nhiu-a

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Sapo 4 years 2020-11-27T01:00:45+00:00 1 Answers 54 views 0

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  1. King
    0
    2020-11-27T01:01:57+00:00
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    `c) cos (3x – π/6) = sin x`

    `<=> cos (3x – π/6) = cos (π/2 – x)`

    `<=>` \(\left[ \begin{array}{l}3x – \dfrac{π}{6} = \dfrac{π}{2} – x + k2π\\3x – \dfrac{π}{6} = -\dfrac{π}{2} + x + k2π\end{array} \right.\) 

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

    `d) cos (5x – π/3) = sin 2x`

    `<=> cos (5x – π/3) = cos (π/2 – 2x)`

    `<=>` \(\left[ \begin{array}{l}5x – \dfrac{π}{3} = \dfrac{π}{2} – 2x + k2π\\5x – \dfrac{π}{3} = -\dfrac{π}{2} + 2x + k2π\end{array} \right.\) 

    `<=>` \(\left[ \begin{array}{l}x = \dfrac{5π}{42} + k\dfrac{2π}{7}\\x = \dfrac{-π}{18} + k\dfrac{2π}{3}\end{array} \right.\) `(k ∈ ZZ)`

    `e) tan (3x – π/4) = cot (x/7)`

    `<=> tan (3x – π/4) = tan (π/2 – x/7)`

    `<=> 3x – π/4 = π/2 – x/7 + kπ`

    `<=> (22x)/7 = (3π)/4 + kπ`

    `<=> x = (21π)/88 + k(7π)/22` `(k ∈ ZZ)`

    `f) cot (x/2 – π/10) = tan (π/5)`

    `<=> cot (x/2 – π/10) = cot (π/2 – π/5)`

    `<=> cot (x/2 – π/10) = cot (3π/10)`

    `<=> x/2 – π/10 = 3π/10 + kπ`

    `<=> x = (4π)/5 + k2π` `(k ∈ ZZ)`

    `b) sin (4x – π/3) = cos 3x`

    `<=> sin (4x – π/3) = sin (π/2 – 3x)`

    `<=>` \(\left[ \begin{array}{l}4x – \dfrac{π}{3} = \dfrac{π}{2} – 3x + k2π\\4x – \dfrac{π}{3} = π – \dfrac{π}{2} + 3x + k2π\end{array} \right.\) 

    `<=>` \(\left[ \begin{array}{l}x = \dfrac{5π}{42} + k\dfrac{2π}{7}\\x = \dfrac{5π}{6} + k2π\end{array} \right.\) `(k ∈ ZZ)`

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