Toán 1. (cosx-2)(cos2x-1)=0 2. Cos3x+cos(x+pi/6)=0 22/07/2021 By Audrey 1. (cosx-2)(cos2x-1)=0 2. Cos3x+cos(x+pi/6)=0
Đáp án: Giải thích các bước giải: (cosx-2)(cos2x-1)=0<=> cosx-2=0 và cos2x-1=0<=>cosx=2(loại) cos2x=1<=> 2x=pi/2 +kpi <=> x=pi/4+kpi/2 cos3x+ cos(x+pi/6)=0<=> cos(x+pi/6)=-cos3x<=> x+pi/6=-3x+k2pi và x+pi/6=3x+k2pi <=> x=pi/24+kpi/2 x=-pi/12+kpi Trả lời
`1) (cos x – 2)(cos 2x – 1) = 0` `<=>` \(\left[ \begin{array}{l}cos x – 2 = 0\\cos 2x – 1 = 0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}cos x = 2 \\cos 2x = 1\end{array} \right.\) `<=> 2x = (π)/2 + kπ` `<=> x = π/4 + k(π)/2` `(k ∈ ZZ)` `2) cos 3x + cos (x + (π)/6) = 0` `<=> cos (x + (π)/6) = -cos 3x` `<=> cos (x + (π)/6) = cos (-3x)` `<=>` \(\left[ \begin{array}{l}x + \frac{π}{6} = -3x + k2π\\x + \frac{π}{6} = 3x + k2π\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}4x = -\frac{π}{6} + k2π\\-2x = – \frac{π}{6} + k2π\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = -\frac{π}{24} + k\frac{π}{2}\\x = \frac{π}{12} – kπ\end{array} \right.\) `(k ∈ ZZ)` Trả lời
Đáp án:
Giải thích các bước giải:
(cosx-2)(cos2x-1)=0<=> cosx-2=0 và cos2x-1=0<=>cosx=2(loại) cos2x=1<=> 2x=pi/2 +kpi
<=> x=pi/4+kpi/2
cos3x+ cos(x+pi/6)=0<=> cos(x+pi/6)=-cos3x<=> x+pi/6=-3x+k2pi và x+pi/6=3x+k2pi
<=> x=pi/24+kpi/2 x=-pi/12+kpi
`1) (cos x – 2)(cos 2x – 1) = 0`
`<=>` \(\left[ \begin{array}{l}cos x – 2 = 0\\cos 2x – 1 = 0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}cos x = 2 \\cos 2x = 1\end{array} \right.\)
`<=> 2x = (π)/2 + kπ`
`<=> x = π/4 + k(π)/2` `(k ∈ ZZ)`
`2) cos 3x + cos (x + (π)/6) = 0`
`<=> cos (x + (π)/6) = -cos 3x`
`<=> cos (x + (π)/6) = cos (-3x)`
`<=>` \(\left[ \begin{array}{l}x + \frac{π}{6} = -3x + k2π\\x + \frac{π}{6} = 3x + k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}4x = -\frac{π}{6} + k2π\\-2x = – \frac{π}{6} + k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = -\frac{π}{24} + k\frac{π}{2}\\x = \frac{π}{12} – kπ\end{array} \right.\) `(k ∈ ZZ)`