Pt $\Leftrightarrow 4cos^22x-(2-\sqrt{3})cos2x-2-\sqrt{3}=sin2x$$\Leftrightarrow 4(cos^22x-1)-(2-\sqrt{3})(cos2x+1)=sin2x$
$\Leftrightarrow (cos2x+1)(4cos2x-2-\sqrt{3})-2sinxcosx=0$
$\Leftrightarrow cos^2x(4cos2x-2-\sqrt{3})-sinxcosx=0$
$\Leftrightarrow cosx(4cosx.cos2x-(2+\sqrt{3})cosx-sinx)=0$
$\Leftrightarrow cosx=0\vee 4cosxcos2x-(2+\sqrt{3})cosx-sinx=0(1)$
$(1)\Leftrightarrow 4cosx(2cos^2x-1)-(2+\sqrt{3})cosx-sinx=0$
$\Leftrightarrow 8cos^3x-(6+\sqrt{3})cosx-sinx=0$
Chia cho $cos^3x$ ta được:
$8-(6+\sqrt{3})(1+tan^2x)-tanx(1+tan^2x)=0$
Từ giải nốt