Đáp án:
$x=\pi/4+k\pi$
Giải thích các bước giải:
$tanx- 2cotx=1$
$<=>\frac{sinx}{cos} - \frac{2cosx}{sinx}=1$
$<=> \frac{sin²x - 2cos²x}{sinx.cosx}=1$
$<=>1-3cos²x=sinx.cosx$
$(đk x\neq k\pi,x \neq\pi/2 +k\pi)$
$<=> 1+cos2x=2sinx.cosx$
$<=> 1 +cos2x=sin2x$
$<=> sin2x -cos2x =1$
$<=>\sqrt[]{2}.sin(2x+ π/4)=1$
$<=> sin(2x+π/4)= 1/√2$
$<=> sin(2x+π/4)= sinπ/4$
<=>\(\left[ \begin{array}{l}2x=k2\pi\\2x=\pi/2+k2\pi\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=k\pi(ko tm dk)\\x=\pi/4+k\pi(tmđk)\end{array} \right.\)
