Đáp án:
`a)` `sin(x^2-2x)=0`
`<=> x^2-2x=kpi` (`kinZZ`)
`<=> x^2-2x+1=kpi+1` (`kinZZ`)
`<=> (x-1)^2=kpi+1` (`kinZZ`)
`<=>` \(\left[ \begin{array}{l}
x-1=\sqrt{k\pi+1}
\\
x-1=-\sqrt{k\pi+1}
\end{array} \right.\) (với `kinZZ`)
`<=>` \(\left[ \begin{array}{l}
x=1+\sqrt{k\pi+1}
\\
x=1-\sqrt{k\pi+1}
\end{array} \right.\) (với `kinZZ`)
Vậy `S={1+sqrt{kpi+1}; 1-sqrt{kpi+1} | kinZZ}`
`b)` `tan(x^2 + 2x + 3) = tan 2`
`<=> x^2 + 2x + 3 = 2 - π + kπ` (`kinZZ`)
`<=> x^2 + 2 x +1= - π + kπ` (`kinZZ`)
`<=> (x+1)^2= -π + kπ` (`kinZZ`)
`<=>` \(\left[ \begin{array}{l}
x+1=\sqrt{-π + kπ}
\\
x+1=-\sqrt{-π + kπ}
\end{array} \right.\) (với `kinZZ`)
`<=>` \(\left[ \begin{array}{l}
x=\sqrt{-π + kπ}-1
\\
x=-\sqrt{-π + kπ}-1
\end{array} \right.\) (với `kinZZ`)
Vậy `S={\sqrt{-π + kπ}-1; -\sqrt{-π + kπ}-1 | kinZZ}`
`c)` `cos2x−cos8x+cos6x=1`
`<=> (cos2x-1)+(cos6x−cos8x)=0`
`<=> 1-2sin^2x-1+2sin({6x+8x}/2)sin({8x-6x}/2)=0`
`<=> -2sin^2x+2sin7x*sinx=0`
`<=> −2sinx(sinx−sin7x)=0`
`<=> 2sinx*2cos4x*sin3x=0`
`<=>`\(\left[ \begin{array}{l}
2\sin x=0
\\
2\cos 4x=0
\\\sin3x=0
\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}
x=k\pi
\\
4x=\frac{\pi}{2}+k\pi
\\3x=k\pi
\end{array} \right.\) (`kinZZ`)
`<=>`\(\left[ \begin{array}{l}
x=k\pi
\\
x=\frac{\pi}{8}+\frac{k\pi}{4}
\\x=\frac{k\pi}{3}
\end{array} \right.\) (`kinZZ`)
Vậy `S={kpi; pi/8+{kpi}/4; {kpi}/3 | kinZZ}`
