[tex]cosx+\sqrt{3}(sin2x+sinx)-2(cos3x+cosx)-2cos^2x+2=0<=>\sqrt{3}(sin2x+sinx)-2cos3x-cosx-2cos^2x+2=0<=>\sqrt{3}sinx(1+2cosx)-cosx(1+2cosx)-8cos^4x+6cosx+2=0<=>\sqrt{3}sinx(1+2cosx)-cosx(1+2cosx)+(1+2cosx)^2(-2cosx+2)=0=>1+2cosx=0[/tex]
hoac : [tex]\sqrt{3}sinx-cosx+(1+2cosx)(-2cosx+2)=0<=>\sqrt{3}sinx-cosx-4cos^2x+2cosx+3=0<=>\sqrt{3}sinx+cosx-4cos^2x+3=0<=>\sqrt{3}sinx+cosx+3sin^2x-cos^2x<=>3sin^2x+\sqrt{3}sinx+\frac{1}{4}-cos^2x+cosx-\frac{1}{4}<=>\left ( \sqrt{3}sinx+\frac{1}{2} \right )^2=\left ( cosx-\frac{1}{2} \right )^2[/tex]
=> 2 th
TH1: [tex]\left ( \sqrt{3}sinx+\frac{1}{2} \right )=\left ( cosx-\frac{1}{2} \right )<=>\sqrt{3}sinx-cosx=-1<=>2sin\left ( x-\frac{\pi}{6} \right )=-1[/tex] ( loai )
TH2: [tex]\left ( \sqrt{3}sinx+\frac{1}{2} \right )=-\left ( cosx-\frac{1}{2} \right )<=>\sqrt{3}sinx+cosx=0=>2sin\left ( x+\frac{\pi}{6} \right )=0=>x+\frac{\pi}{6}=k\pi=>x=k\pi-\frac{\pi}{6}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
