Toán Lớp 11: Giải phương trình:
1+cot2x=(1-cos2x)/(sin^2 2x)
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Toán Lớp 11: Giải phương trình:
1+cot2x=(1-cos2x)/(sin^2 2x)
Comments ( 2 )
$1+\cot \left(2x\right)=\dfrac{1-\cos \left(2x\right)}{\sin ^2\left(2x\right)}$
$⇔1+\cot \left(2x\right)-\dfrac{1-\cos \left(2x\right)}{\sin ^2\left(2x\right)}=0$
$⇔\dfrac{\sin ^2\left(2x\right)+\sin ^2\left(2x\right)\cot \left(2x\right)-1+\cos \left(2x\right)}{\sin ^2\left(2x\right)}=0$
$⇔\sin ^2\left(2x\right)+\sin ^2\left(2x\right)\cot \left(2x\right)-1+\cos \left(2x\right)=0$
$⇔\cos \left(2x\right)-\cos ^2\left(2x\right)+\left(1-\cos ^2\left(2x\right)\right)\cot \left(2x\right)=0$
$⇔\left(1-\cos \left(2x\right)\right)\left(\cos \left(2x\right)+\cot \left(2x\right)\left(1+\cos \left(2x\right)\right)\right)=0$
$⇔\begin{cases} 1-\cos \left(2x\right)=0\\\cos \left(2x\right)+\cot \left(2x\right)\left(1+\cos \left(2x\right)\right)=0 \end{cases}$ $⇔\begin{cases} x=k\pi \\x=\dfrac{\pi }{4}+k\pi ,\:x=k\pi +\dfrac{3\pi }{4} \end{cases}$
$⇔\begin{cases} x=\dfrac{\pi }{4}+k\pi \\x=k\pi +\dfrac{3\pi }{4} \end{cases}$
Giải đáp:
Lời giải và giải thích chi tiết:
ĐKXĐ $ : sin2x \neq 0 <=> x \neq \dfrac{k\pi}{2}$
$ PT <=> sin^{2}2x + sin2xcos2x = 1 – cos2x$
$ <=> sin2xcos2x – (1 – sin^{2}2x) + cos2x = 0$
$ <=> sin2xcos2x – cos^{2}2x + cos2x =‘0$
$ <=> cos2x(sin2x – cos2x + 1) = 0$
– TH1 $ : cos2x = 0 <=> x = (2k + 1)\dfrac{\pi}{4} $
– TH2 $: cos2x – sin2x = 1$
$ => sin^{2}2x + cos^{2}2x – 2sin2xcos2x = 1$
$ <=> sin2xcos2x = 0 <=> cos2x = 0 $ (vì $ sin2x \neq 0)$
$ <=> 2x = (2k + 1)\dfrac{\pi}{2} <=> x = (2k + 1)\dfrac{\pi}{4}$