Giải phương trình: `1+cot2x=(1-cos2x)/(sin^2 2x)`

$1+\cot \left(2x\right)=\dfrac{1-\cos \left(2x\right)}{\sin ^2\left(2x\right)}$

$⇔1+\cot \left(2x\right)-\dfrac{1-\cos \left(2x\right)}{\sin ^2\left(2x\right)}=0$

$⇔\dfrac{\sin ^2\left(2x\right)+\sin ^2\left(2x\right)\cot \left(2x\right)-1+\cos \left(2x\right)}{\sin ^2\left(2x\right)}=0$

$⇔\sin ^2\left(2x\right)+\sin ^2\left(2x\right)\cot \left(2x\right)-1+\cos \left(2x\right)=0$

$⇔\cos \left(2x\right)-\cos ^2\left(2x\right)+\left(1-\cos ^2\left(2x\right)\right)\cot \left(2x\right)=0$

$⇔\left(1-\cos \left(2x\right)\right)\left(\cos \left(2x\right)+\cot \left(2x\right)\left(1+\cos \left(2x\right)\right)\right)=0$

$⇔\begin{cases} 1-\cos \left(2x\right)=0\\\cos \left(2x\right)+\cot \left(2x\right)\left(1+\cos \left(2x\right)\right)=0 \end{cases}$ $⇔\begin{cases} x=k\pi \\x=\dfrac{\pi }{4}+k\pi ,\:x=k\pi +\dfrac{3\pi }{4} \end{cases}$

$⇔\begin{cases} x=\dfrac{\pi }{4}+k\pi \\x=k\pi +\dfrac{3\pi }{4} \end{cases}$

Đáp án:

Giải thích các bước giải:

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 $sin2x \neq 0)$
$<=> 2x  = (2k + 1)\dfrac{\pi}{2} <=> x = (2k + 1)\dfrac{\pi}{4}$