lượng giác

N

nguyenbahiep1

đặt
x = 3u

[TEX]cos(4u) = cos^2 (3u) \\ 2.cos^2 2u -1 = (4cos^3 u -3cosu)^2 \\ 2.(2cos^2u -1)^2 -1 = 16cos^6 u -24cos^4u + 9cos^2u \\ 8cos^4u -8cos^2u +1 = 16cos^6 u -24cos^4u + 9cos^2u \\ 16.cos^6u -32cos^4u +17cos^2u -1 = 0 \\ t = cos^2u \\ 16.t^3 -32.t^2 +17.t -1 = 0 \\ t = 1 \Rightarrow cos^2u = 1 \\ t = \frac{8-\sqrt{48}}{16} = \frac{2-\sqrt{3}}{4} \\ t = \frac{8+\sqrt{48}}{16}= \frac{2+\sqrt{3}}{4}[/TEX]

phần còn lại là của bạn nhé
 
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N

newstarinsky

$cos(x+\dfrac{x}{3})=cos^2x\\
\Leftrightarrow cosx.cos\dfrac{x}{3}-sinx.sin\dfrac{x}{3}=cos^2x\\
\Leftrightarrow cosx(cos\dfrac{x}{3}-cosx)-sinx.sin\dfrac{x}{3}=0\\
\Leftrightarrow 2cosx.sin\dfrac{2x}{3}.sin\dfrac{x}{3}-sinx.sin\dfrac{x}{3}=0\\
\Leftrightarrow sin\dfrac{x}{3}(4cosx.sin\dfrac{x}{3}.cos\dfrac{x}{3}-3sin\dfrac{x}{3}+4sin^3\dfrac{x}{3})=0\\
\Leftrightarrow sin^2\dfrac{x}{3}[4(4cos^3\dfrac{x}{3}-3cos\dfrac{x}{3}).cos\dfrac{x}{3}-3+4(1-cos^2\dfrac{x}{3})]=0\\
\Leftrightarrow sin^2\dfrac{x}{3}(16cos^4\dfrac{x}{3}-16cos^2\dfrac{x}{3}+1)=0$
OK nhé
 
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