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N

nguyenbahiep1

(tanx.cos3x-2cos2x+1)/1-2sinx=căn 3 (sin2x+cosx)
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[laTEX]TXD: 1-2sinx \not = 0 \\ \\ cosx \not = 0 \\ \\ sinx.cos3x - 2cos2x.cosx + cosx = \sqrt{3}cos^2x(1-4sin^2x) \\ \\ \frac{1}{2}.(sin4x - sin2x) - cos3x - cosx+cosx = \sqrt{3}cos^2x(1-4sin^2x) \\ \\ 2sin2x.cos2x - sin2x - 2(4cos^3x-3cosx) = 2\sqrt{3}cos^2x(1-4sin^2x) \\ \\ TH_1: cosx = 0 \\ \\ TH_2: 2sinx.cos2x - sinx - (4cos^2x-3) = \sqrt{3}cosx(1-4sin^2x) \\ \\ 2sinx(1-2sin^2x) - sinx - (1-4sin^2x) = \sqrt{3}cosx(1-4sin^2x) \\ \\ sinx(1-4sin^2x) - (1-4sin^2x) = \sqrt{3}cosx(1-4sin^2x) \\ \\ \Rightarrow 1-4sin^2x = 0 \\ \\ sinx - \sqrt{3}cosx = 1 [/laTEX]
 
C

connhikhuc

luavangroi_97@yahoo.com.vn said:
(tanx.cos3x-2cos2x+1)/1-2sinx=căn 3 (sin2x+cosx);)
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pt \Leftrightarrow [TEX][\frac{sinx}{cosx}.cos3x-(2cos2x-1)]/1-2sinx=\sqrt[]{3}(sin2x+cosx)[/TEX]

\Leftrightarrow [TEX][\frac{sin4x-sin2x}{2cosx}-(2cosx-1)]/1-2sinx=\sqrt[]{3}cosx(2sinx+1)[/TEX]

\Leftrightarrow [TEX][sin2x\frac{2cos2x-1}{2cosx}-(2cosx-1)]/1-2sinx=\sqrt[]{3}cosx(2sinx+1)[/TEX]

\Leftrightarrow [TEX]\frac{2cosx(2cos2x-1)(sinx-1)}{2cosx(1-2sinx)}=\sqrt[]{3}cosx(2sinx+1)[/TEX]

\Leftrightarrow [TEX]\frac{2cosx(1-4sin^2 x)(sinx-1)}{2cosx(1-2sinx)}=\sqrt[]{3}cosx(2sinx+1)[/TEX]

\Leftrightarrow [TEX]\frac{2cosx(1-2sinx)(1+2sinx)(sinx-1)}{2cosx(1-2sinx)}=\sqrt[]{3}cosx(2sinx+1)[/TEX]

nhom lai va giai , lay dk
 
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