Toán Lớp 11: Giải các phương trình sau:
a) $\sin(x^2-2x)=0$
b) $\tan(x^2+2x+3)=\tan 2$
c) $\cos 2x-\cos 8x+\cos 6x=1$
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Toán Lớp 11: Giải các phương trình sau:
a) $\sin(x^2-2x)=0$
b) $\tan(x^2+2x+3)=\tan 2$
c) $\cos 2x-\cos 8x+\cos 6x=1$
Comments ( 2 )
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}