b). Sinx + sin3x + sin4x = 0 c) 2cosx.cos2x = 1 + cos2x + cos3x 23/07/2021 Bởi Vivian b). Sinx + sin3x + sin4x = 0 c) 2cosx.cos2x = 1 + cos2x + cos3x
a) `sinx+sin3x+sin4x = 0` `<=> 2sin2xcosx + 2sin2xcos2x =0` `<=> sin2x(cosx + cos2x)=0` `<=>` \(\left[ \begin{array}{l}sin2x=0\\cosx=cos2x\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}2x=kπ\\x= \pm 2x + k2π\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{kπ}{2}\\x=k2π\\x=\dfrac{k2π}{3}\end{array} \right.\) Vậy PT có 2 họ nghiệm : `x = (kπ)/2` và `x = (k2π)/3` b) `2cosx . cos2x = 1 + cos2x + cos3x` `<=> cosx + cos3x = 1 + 2cos^2x – 1 + cos3x` `<=> 2cos^2x – cosx =0` `<=> cosx(2cosx-1) =0` `<=>` \(\left[ \begin{array}{l}cosx=0\\cosx=\dfrac{1}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{π}{2}+kπ\\x=\pm \dfrac{π}{3}+k2π\end{array} \right.\) Vậy PT có 2 họ nghiệm: `x=π/2 + kπ` và `x = π/3 + kπ` Bình luận
a) `sinx+sin3x+sin4x = 0`
`<=> 2sin2xcosx + 2sin2xcos2x =0`
`<=> sin2x(cosx + cos2x)=0`
`<=>` \(\left[ \begin{array}{l}sin2x=0\\cosx=cos2x\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=kπ\\x= \pm 2x + k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{kπ}{2}\\x=k2π\\x=\dfrac{k2π}{3}\end{array} \right.\)
Vậy PT có 2 họ nghiệm : `x = (kπ)/2` và `x = (k2π)/3`
b) `2cosx . cos2x = 1 + cos2x + cos3x`
`<=> cosx + cos3x = 1 + 2cos^2x – 1 + cos3x`
`<=> 2cos^2x – cosx =0`
`<=> cosx(2cosx-1) =0`
`<=>` \(\left[ \begin{array}{l}cosx=0\\cosx=\dfrac{1}{2}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{π}{2}+kπ\\x=\pm \dfrac{π}{3}+k2π\end{array} \right.\)
Vậy PT có 2 họ nghiệm: `x=π/2 + kπ` và `x = π/3 + kπ`