Hỡi thế gian ai là người giỏi toán,

Giải dùm tôi bài toán tam sư !

Giải thích các bước giải:

ĐKXĐ: $x\notin\{\dfrac{11}6\pi+k2\pi, \dfrac76\pi+k2\pi, \dfrac12\pi+k2\pi\}$

Ta có:
$\dfrac{(1-2\sin x)\cos x}{(1+2\sin x)(1-\sin x)}=\sqrt3$

$\to (1-2\sin x)\cos x=\sqrt3(1+2\sin x)(1-\sin x)$

$\to \cos x-2\sin x\cos x= \sqrt3(1+2\sin x-\sin x+2\sin^2x)$

$\to \cos x-2\sin x\cos x= \sqrt3(\sin x+1+2\sin^2x)$

$\to \cos x-\sin2x= \sqrt3(\sin x+\cos2x)$

$\to \cos x-\sqrt3\sin x=\sin2x+\sqrt3\cos2x$

$\to \dfrac12\cos x-\dfrac{\sqrt3}2\sin x=\dfrac12\sin2x+\dfrac{\sqrt3}{2}\cos2x$

$\to \cos(x+\dfrac13\pi)=\cos(2x-\dfrac16\pi)$

$\to x+\dfrac13\pi=2x-\dfrac16\pi+k2\pi\to x=\dfrac12\pi-2k\pi$

Hoặc $x+\dfrac13\pi=-(2x-\dfrac16\pi)+k2\pi\to x=-\dfrac1{18}\pi+\dfrac3k\pi$

`((1-2sinx).cosx)/((1+2sinx)(1-sinx)) =sqrt3`

`<=>(1-2sinx).cosx=sqrt3(1+2sinx)(1-sinx)`

`<=>cosx-2sinx cosx=sqrt3(1-sinx+2sinx-2sin^2x)`

`<=>cosx-2sinx cosx=sqrt3(1+sinx-2sin^2x)`

`<=>cosx-2sinxcosx=sqrt3(1+sinx-2.(1-cos2x)/2)`

`<=>cosx-2sinxcosx=sqrt3(sinx+cos2x)`

`<=>cosx-sin2x=sqrt3 sinx +sqrt3 cos2x`

`<=>cosx -sqrt3 sinx =sin2x+sqrt3 cos 2x `

`<=>1/2( cosx -sqrt3 sinx)=1/2 (sin 2x+sqrt3 cos2x)`

`<=>cos(x+pi/3)=cos(2x-pi/6)`

`<=>``[(2x-pi/6=x+pi/3+k2pi),(2x-pi/6=pi/3 +x +k2pi):}`

`<=>[(x=pi/2+k2pi),(x=(k2pi)/3 -pi/18):} (k in ZZ)`