Toán Giải phương trình ( sin3x + cos3x )^2 + căn3 cos6x = 2 07/07/2021 By Savannah Giải phương trình ( sin3x + cos3x )^2 + căn3 cos6x = 2
Đáp án: \(\left[ \begin{array}{l}x=-\frac{\pi}{36}+k\frac{\pi}{3}\\x=\frac{\pi}{12}+k\frac{\pi}{3}\end{array} \right.\) k ∈ Z Giải thích các bước giải: (sin3x+cos3x)² + $\sqrt{3}$ cos6x = 2 <=> 1 + 2sin3xcos3x + $\sqrt{3}$ cos6x = 2 <=> sin6x + $\sqrt{3}$ cos6x = 1 <=> sin(6x+$\frac{\pi}{3}$ ) = $\frac{1}{2}$ <=> \(\left[ \begin{array}{l}x=-\frac{\pi}{36}+k\frac{\pi}{3}\\x=\frac{\pi}{12}+k\frac{\pi}{3}\end{array} \right.\) k ∈ Z Trả lời
`(sin 3x + cos 3x)^2 + sqrt{3}cos 6x = 2` `<=> sin^2 3x + cos^2 3x + 2sin 3x.cos 3x + sqrt{3}cos 6x = 2` `<=> 1 + 2sin 3x.cos 3x + sqrt{3}cos 6x = 2` `<=> sin 6x + sqrt{3}cos 6x = 1` `<=> 2.sin (6x + π/3) = 1` `<=> sin (6x + π/3) = sin (π/6)` `<=>` \(\left[ \begin{array}{l}6x + \dfrac{π}{3} = \dfrac{π}{6} + k2π\\6x + \dfrac{π}{3} = \dfrac{5π}{6} + k2π\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = -\dfrac{π}{36} + k\dfrac{π}{3}\\x = \dfrac{π}{2} + k\dfrac{π}{3}\end{array} \right.\) `(k ∈ ZZ)` Trả lời
Đáp án:
\(\left[ \begin{array}{l}x=-\frac{\pi}{36}+k\frac{\pi}{3}\\x=\frac{\pi}{12}+k\frac{\pi}{3}\end{array} \right.\) k ∈ Z
Giải thích các bước giải:
(sin3x+cos3x)² + $\sqrt{3}$ cos6x = 2
<=> 1 + 2sin3xcos3x + $\sqrt{3}$ cos6x = 2
<=> sin6x + $\sqrt{3}$ cos6x = 1
<=> sin(6x+$\frac{\pi}{3}$ ) = $\frac{1}{2}$
<=> \(\left[ \begin{array}{l}x=-\frac{\pi}{36}+k\frac{\pi}{3}\\x=\frac{\pi}{12}+k\frac{\pi}{3}\end{array} \right.\) k ∈ Z
`(sin 3x + cos 3x)^2 + sqrt{3}cos 6x = 2`
`<=> sin^2 3x + cos^2 3x + 2sin 3x.cos 3x + sqrt{3}cos 6x = 2`
`<=> 1 + 2sin 3x.cos 3x + sqrt{3}cos 6x = 2`
`<=> sin 6x + sqrt{3}cos 6x = 1`
`<=> 2.sin (6x + π/3) = 1`
`<=> sin (6x + π/3) = sin (π/6)`
`<=>` \(\left[ \begin{array}{l}6x + \dfrac{π}{3} = \dfrac{π}{6} + k2π\\6x + \dfrac{π}{3} = \dfrac{5π}{6} + k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = -\dfrac{π}{36} + k\dfrac{π}{3}\\x = \dfrac{π}{2} + k\dfrac{π}{3}\end{array} \right.\) `(k ∈ ZZ)`