Giải các phương trình sau :
1.sin(2x – 1)= -sinx
2.Cot( -x +2π/9)= cot(-3x + π/2)
3.Sin(- x + π/3)= sin(x + 3π/8)
Giải các phương trình sau :
1.sin(2x – 1)= -sinx
2.Cot( -x +2π/9)= cot(-3x + π/2)
3.Sin(- x + π/3)= sin(x + 3π/8)
Đáp án:
Giải thích các bước giải:
1. `sin(2x – 1)= -sinx`
`⇔ sin(2x-1)=sin(-x)`
`⇔` \(\left[ \begin{array}{l}2x-1=-x+k2\pi\ (k \in \mathbb{Z})\\2x-1=\pi+x+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{1}{3}+\dfrac{k2\pi}{3}\ (k \in \mathbb{Z})\\x=1+\pi+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
2. `Cot( -x +(2π)/9)= cot(-3x + π/2)`
`⇔ -x+\frac{2\pi}{9}=-3x+\frac{\pi}{2}+k\pi\ (k \in \mathbb{Z})`
`⇔ 2x=\frac{5}{18}\pi+k\pi\ (k \in \mathbb{Z})`
`⇔ x=\frac{5}{36}\pi+\frac{k\pi}{2}\ (k \in \mathbb{Z})`
3. `Sin(- x + π/3)= sin(x + (3π)/8)`
`⇔` \(\left[ \begin{array}{l}-x+\dfrac{\pi}{3}=x+\dfrac{3\pi}{8}+k2\pi\ (k \in \mathbb{Z})\\-x+\dfrac{\pi}{3}=\pi-x-\dfrac{3\pi}{8}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{1}{48}\pi-k\pi\ (k \in \mathbb{Z})\\0x=\dfrac{7}{24}\pi+k2\pi\ (vô\ lí)\end{array} \right.\)