Chủ đề này có 2 trả lời

[cityhuntervp]

$1+3tanx=2sin2x$

[AnnieSally]

$PT\Leftrightarrow 1+3tanx=2.\dfrac{2tanx}{1+tan^{2}x}\Leftrightarrow (1+3tanx)(1+tan^{2}x)=4tanx\Leftrightarrow (tanx+1)(3tan^{2}x-2tanx+1)=0\Leftrightarrow tanx+1=0\Leftrightarrow tanx=-1\Leftrightarrow x=\dfrac{-\pi}{4}+k\pi$

[ntsondn98]

1+3tanx=2sin2x

<=>$1+3\frac{sinx}{cosx}=4sinxcosx$ (1)

 

dk $x\neq \frac{\Pi }{2}+k\Pi$

 

(1)<=>$cosx+3sinx-4sinxcos^{2}x=0$

<=>$cosx(sin^{2}+cos^{2}x)+3sinx(sin^{2}x+cos^{2}x)-4sinxcos^{2}x=0$

<=>$sin^{2}xcosx+cos^{3}x+3sin^{3}x+3sinxcos^{2}x-4sinxcos^{2}x=0$

<=>$3sin^{3}x+sin^{2}xcosx-sinxcos^{2}x+cos^{3}x=0$

<=>$3tan^{3}x+tan^{2}x-tanx+1=0$

<=>tanx=-1

<=>$x=\frac{-\Pi }{4}+k\Pi$