Toán Lớp 11: giải pt
sin2x=2sinxcos(x-pi/3)
4sin^2x=1
2sin^2(2x+pi/3)=2sinxcosx +1
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Toán Lớp 11: giải pt
sin2x=2sinxcos(x-pi/3)
4sin^2x=1
2sin^2(2x+pi/3)=2sinxcosx +1
Comments ( 1 )
sin2x=2sinxcos(x-π/3)
<=>sin2x-2sinxsin(x+π/6)=0
<=>2sin(π/6-x)sinx=0
<=>sin(π/6-x)sinx=0
<=> \(\left[ \begin{array}{l}
\sin(\frac{π}{6}-x)=0\\
\sin x=0
\end{array} \right.\)
<=> \(\left[ \begin{array}{l}
\frac{π}{6}-x=πn
\\
x = π n
\end{array} \right.\) (với n\inZ)
<=> \(\left[ \begin{array}{l}
x=\frac{π}{6}-πn
\\
x = π n
\end{array} \right.\) (với n\inZ)
Vậy S={π/6 – π n; π n | n in Z}
$\\$
4sin^2x=1
<=> sin^2x=1/4
<=> \(\left[ \begin{array}{l}
\sin x = \frac{1}{2}
\\
\sin x = \frac{-1}{2}
\end{array} \right.\)
<=> \(\left[ \begin{array}{l}
x = \frac{5 π}{6}+2 π n
\\
x = \frac{π}{6}+2 π n
\\
x=\frac{7 π}{6}+2 π n
\\
x=\frac{11 π}{6}+2 π n
\end{array} \right.\) (với n\inZ)
Vậy S={(5 π)/6+2 π n; π/6+2 π n; (7 π)/6+2 π n; (11 π)/6+2 π n | n\in Z}
$\\$
2 sin^2(2 x + π/3) = 2 sinx cosx + 1
<=> -1 + 2 cos^2(-2 x + π/6) – 2 sinx cosx = 0
<=> 2 cos(3 x + π/12) sin(x + π/12) = 0
<=> cos(3 x + π/12) sin(x + π/12) = 0
<=> \(\left[ \begin{array}{l}
\cos(3 x + \frac{π}{12}) = 0\\
\sin(x + \frac{π}{12}) = 0
\end{array} \right.\)
<=> \(\left[ \begin{array}{l}
3 x + \frac{π}{12} = \frac{π}{2}+π n
\\
x + \frac{π}{12} = π n
\end{array} \right.\) (với n\inZ)
<=> \(\left[ \begin{array}{l}
3 x= \frac{5π}{12}+π n
\\
x = π n – \frac{π}{12}
\end{array} \right.\) (với n\inZ)
<=> \(\left[ \begin{array}{l}
x= \frac{5π}{36}+\frac{πn}{3}
\\
x = π n – \frac{π}{12}
\end{array} \right.\) (với n\inZ)
Vậy S={(5 π)/36+(π n)/3; π n – π/12 | n in Z}