3sin3x- $\sqrt[]{3}$ cos9x =1+4 $sin^{3}$3x 28/08/2021 Bởi Daisy 3sin3x- $\sqrt[]{3}$ cos9x =1+4 $sin^{3}$3x
Đáp án: \(\left[ \begin{array}{l}x=\dfrac{π}{18}+k \dfrac{2π}{9}\\x=\dfrac{7π}{54} + k \dfrac{2π}{9}\end{array} \right.\) Giải thích các bước giải: ` 3sin3x – \sqrt3 cos9x = 1 +4sin^3 3x` `<=> (3sin3x – 4sin^3 3x) – \sqrt3 cos 9x = 1` `<=> sin9x – \sqrt3 cos9x = 1` `<=> 1/2 sin9x – (\sqrt3)/2 cos 9x = 1/2` `<=> sin9x . cos π/3 – cos 9x . sin π/3 = sin π/6` `<=> sin ( 9x – π/3) = sin π/6` `<=>` \(\left[ \begin{array}{l}9x- \dfrac{π}{3} = \dfrac{π}{6} + k2π\\9x-\dfrac{π}{3}=π -\dfrac{π}{6} + k2π\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{π}{18}+k \dfrac{2π}{9}\\x=\dfrac{7π}{54} + k \dfrac{2π}{9}\end{array} \right.\) Bình luận
Đáp án: \(\left[ \begin{array}{l}x=\dfrac{π}{18}+k \dfrac{2π}{9}\\x=\dfrac{7π}{54} + k \dfrac{2π}{9}\end{array} \right.\)
Giải thích các bước giải:
` 3sin3x – \sqrt3 cos9x = 1 +4sin^3 3x`
`<=> (3sin3x – 4sin^3 3x) – \sqrt3 cos 9x = 1`
`<=> sin9x – \sqrt3 cos9x = 1`
`<=> 1/2 sin9x – (\sqrt3)/2 cos 9x = 1/2`
`<=> sin9x . cos π/3 – cos 9x . sin π/3 = sin π/6`
`<=> sin ( 9x – π/3) = sin π/6`
`<=>` \(\left[ \begin{array}{l}9x- \dfrac{π}{3} = \dfrac{π}{6} + k2π\\9x-\dfrac{π}{3}=π -\dfrac{π}{6} + k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{π}{18}+k \dfrac{2π}{9}\\x=\dfrac{7π}{54} + k \dfrac{2π}{9}\end{array} \right.\)
Đáp án:
Giải thích các bước giải: